Some introductory Topology questions

[micromass]

My book gives a definition in an appendix that seems to match what you're saying.

Unfortunately, I'm having a hard time grasping the idea (the set you've wrote, I can read, but I can't "see it."). It is the set of all possible sequences of any length in R? Is that like a power set of R, but where order matters for the elements? Or am I way off?

It is the set of sequences of infinite length.

Put another way, it is the set of all functions $f:\mathbb{N}\rightarrow \mathbb{R}$.

 

[1MileCrash]

OK, I've been thinking about it for a while.

From what I gather from that function notation you've used, it is essentially a declaration that these elements are also of countably infinite length, since by noting N as the domain, it must also have a one-to-one correspondence with the natural numbers. Right?

 

[micromass]

OK, I've been thinking about it for a while.

From what I gather from that function notation you've used, it is essentially a declaration that these elements are also of countably infinite length, since by noting N as the domain, it must also have a one-to-one correspondence with the natural numbers. Right?

Not necessarily a one-to-one correspondence. For example

$$f:\mathbb{N}\rightarrow \mathbb{R}: n\rightarrow 1$$

is perfectly valid. This corresponds to a sequence $(x_n)_n$ with $x_n = 1$ for all $n$.

 

[1MileCrash]

Not necessarily a one-to-one correspondence. For example

$$f:\mathbb{N}\rightarrow \mathbb{R}: n\rightarrow 1$$

is perfectly valid. This corresponds to a sequence $(x_n)_n$ with $x_n = 1$ for all $n$.

Errm, I must not have phrased it well. When I said "countably infinite length," I am speaking of cardinality. I don't mean that any element of R^omega is a tuple where every number is different (because then every element of R^omega would be the same except for order)

$$f:\mathbb{N}\rightarrow \mathbb{R}: n\rightarrow 1$$

Tells me that the resulting ordered set is a countable one (and thus has a one to one correspondence with the natural numbers, in terms of cardinality, not any values) because our domain is N. If our domain is N then our tuple has the same cardinality as N, it doesn't matter that values are repeated in the tuple.

Essentially, I'm lead to believe that all R^(omega) is is just an "countably infinite dimensional space." If R^3 is the set of all possible 3-tuples, and R^omega is the set of all possible "countably infinite-tuples" then R^omega doesn't seem to be very special.

 

[micromass]

OK, I agree with you then.

 

[1MileCrash]

OK, I agree with you then.

Alright. So is looking at R^omega as just being exactly like any other dimensional space R^n except that now, n is no natural number but is countable infinity, an OK way to view it? Or is it more complicated than that?

Thanks again for your help and patience.

 

[micromass]

Alright. So is looking at R^omega as just being exactly like any other dimensional space R^n except that now, n is no natural number but is countable infinity, an OK way to view it? Or is it more complicated than that?

Thanks again for your help and patience.

That's one way to think of it, yes. As you know, $\mathbb{R}^n$ is a vector space with finite dimensions. The space $\mathbb{R}^\omega$ is a vector space too, so it acts in the same way of $\mathbb{R}^n$. The only difference is that $\mathbb{R}^\omega$ is an infinite dimensional vector space, so that might cause some differences.

You got to be aware that $\mathbb{R}^n$ and $\mathbb{R}^\omega$ often behave very differently, especially in topology. So it's not because it holds for $\mathbb{R}^n$, that it holds for $\mathbb{R}^\omega$. But I do think your intuition about $\mathbb{R}^\omega$ is correct.

 

[1MileCrash]

That's one way to think of it, yes. As you know, $\mathbb{R}^n$ is a vector space with finite dimensions. The space $\mathbb{R}^\omega$ is a vector space too, so it acts in the same way of $\mathbb{R}^n$. The only difference is that $\mathbb{R}^\omega$ is an infinite dimensional vector space, so that might cause some differences.

You got to be aware that $\mathbb{R}^n$ and $\mathbb{R}^\omega$ often behave very differently, especially in topology. So it's not because it holds for $\mathbb{R}^n$, that it holds for $\mathbb{R}^\omega$. But I do think your intuition about $\mathbb{R}^\omega$ is correct.

OK, I do think I can see why some things will be different. I can't imagine a vector of infinite length in any other space, but a vector in R^omega can be of infinite length easily. Then maybe that makes certain things undefined, like a difference between two distance functions on R^omega, and maybe that effects certain theorems. Just thinking out loud!

 

[micromass]

OK, I do think I can see why some things will be different. I can't imagine a vector of infinite length in any other space, but a vector in R^omega can be of infinite length easily. Then maybe that makes certain things undefined, like a difference between two distance functions on R^omega, and maybe that effects certain theorems. Just thinking out loud!

It's good that you mention distance functions (metrics). This is one difference that you should be aware of.

In things like $\mathbb{R}^2$, we have a distance function as follows

$$d((x,y),(x',y')) = \sqrt{(x-x')^2 + (y-y')^2}$$

You showed in this thread, that we can generalize that on $\mathbb{R}^n$ as

$$d((x_1,...,x_n),(y_1,...,y_n)) = \sqrt{\sum_{i=1}^n (x_i - y_i)^2}$$

However, this generalization does not work for $\mathbb{R}^\omega$. The natural thing to do is of course

$$d((x_1,x_2,...),(y_1,y_2,...)) = \sqrt{\sum_{i=1}^{+\infty} (x_i - y_i)^2}$$

But then we are dealing with an infinite series on the RHS. And this might diverge. So the natural metric that we want may not be defined. This is one possible difference.

That said, it is possible to put some metric on $\mathbb{R}^\omega$, but it's not a very nice one.

 

[1MileCrash]

Of course.. a metric is f:XxX->R, but if the function could possibly diverge, then the codomain is not R, so that function can't be a metric.

 

[1MileCrash]

I'm baack... I took a break from topology to learn a lot of linear algebra. I've moved on from metrics to the definitions of topologies and topological spaces.

It took a while for me to comprehend the metric and topology relationship, but I think I'm starting to understand it.

My book says that the discrete metric generates the discrete topology, which is the collection of the empty set and the set itself. So, I infer that a topology is a collection of sets such that if you take any two elements from a member of the topology, and run them through the metric, you get some value? Is that right?

That's the only way I can make the connection between the discrete metric and the discrete topology. Since for any two non-equal values for the discrete metric, the result is 1, it makes sense to me that this is the defining quality that generates the elements of the topology.I do not mean to infer that all topologies have an associated metric, I understand that that is not the case, because a "metrizable" topology is a special kind. The way I have begun to understand it is that a metric outputs values of "distance" from elements of a set, and that we can make a collection of subsets of X such that the distinction between members of this collection is the distance value returned by the metric for its members. We can call this a topology. We can then look at the properties of the topology directly and call any collection of subsets of X a topology if they also have these properties, even if they have no metric associated with them.

Is this remotely correct, or am I way in left field?

 

[micromass]

Sorry, but I really don't get what you're trying to say here:

I infer that a topology is a collection of sets such that if you take any two elements from a member of the topology, and run them through the metric, you get some value?

we can make a collection of subsets of X such that the distinction between members of this collection is the distance value returned by the metric for its members.

I don't even know what you mean, so I can't interpret whether you have the right intuition or not.

 

[1MileCrash]

Sorry, let me try to say that in a better way, I typed that in a rush:

My book talks about topologies that are "generated" from metrics, but it doesn't really explain how that happens (how to I "generate" a collection of subsets of X (topology) from a metric on X?)

But, since it listed the examples of topologies that are generated from metrics, I can try to put it together myself.

The simplest example they gave is that the discrete metric (which is the metric d(x,y) such that d = 1 if x=/=y and d = 0 if x = y) generates a topology called the discrete topology on X which is the collection of the following subsets of X: { ∅, X }.

So, knowing that the discrete metric generates the discrete topology, I'm trying to find a special link between the two, so that I can see how a metric generates a topology.

The only connection that I can make is that if I take any two elements of X, I get d=1, so all elements of X share that property.

Then, I wonder if that is the property that separates the collection of subsets of a topology. In other words, is the discrete topology composed of only the sets ∅ and X because all elements of X share the property that d = 1 for any two, and there is no other possibility (and thus members of the topology) for the discrete metric?Thanks again

 

[jbunniii]

The simplest example they gave is that the discrete metric (which is the metric d(x,y) such that d = 1 if x=/=y and d = 0 if x = y) generates a topology called the discrete topology on X which is the collection of the following subsets of X: { ∅, X }.

That is the indiscrete topology (also known as the trivial topology), not the discrete topology. The discrete topology is the power set of X, i.e. the collection of all subsets of X.

The discrete metric generates the discrete topology because every singleton in X is an open set (for example, the neighborhood of radius 1/2 around the point x consists of just the point x). Also, any union of open sets is an open set, and every subset of X can be expressed as a union of singletons. So every subset of X is open.

 

[micromass]

Pretty weird that they don't explain what it means for a topology to be generated by a metic.

Anyway, given a metric space $(X,d)$, we can define the open ball centered in $x\in X$ with radius $r>0$ as

$$B(x,r) := \{y\in X~\vert~d(x,y) < r\}$$

Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$. I assume these are all concepts you've already seen??

Anyway, take your metric space $(X,d)$. The topology generated by the metric space is simply the collection of all open sets.

Also, the discrete metric on a set $X$ induces the topology $\mathcal{P}(X)$ (the power set). So any subset of $X$ is in the topology. In your previous post, you seemed to imply that the topology was $\{\emptyset, X\}$, this is not the case. The latter topology is the indiscrete or trivial topology.

 

[1MileCrash]

Yes, I looked back and had the discrete and indiscreet topologies mixed up.

Pretty weird that they don't explain what it means for a topology to be generated by a metic.

Anyway, given a metric space $(X,d)$, we can define the open ball centered in $x\in X$ with radius $r>0$ as

$$B(x,r) := \{y\in X~\vert~d(x,y) < r\}$$

Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$. I assume these are all concepts you've already seen??

Yes, my book has defined open balls. From what I can tell, it is just a set of y such that the "distance" between x and y is less than r.

The definition of open sets you gave was also given, but it's clear now that I should have spent more time on them, I'm having trouble getting a grasp over it.

It says that G is open if for any x in G, there is an r > 0 such that B(x,r) is a subset of G. So if for any x, there is an r such that the set of y such that the "distance" between x and y is less than r is a subset of G.

That is really confusing to me, I can't really make sense of it. I already understand open sets as a layman, in that it just "doesn't contain its boundary points" but I can't comprehend it in terms of open balls very well.

So, I can pick any element x of G,
and can find an r such that a set of y such that d(x,y) < r
is a subset of G.

Never before have I been able to type something and not "see it" or understand really what it implies at such a high degree. I really have no idea what this says about G.

Anyway, take your metric space $(X,d)$. The topology generated by the metric space is simply the collection of all open sets.

Ok

How would one do this?

Consider all subsets of X and then see it were true that "I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G." for each one?

Like, for the set X={1,2,3}
{{1}, {2}, {1,2},{2,3},∅, X}}

is a topology. So each element is an "open set."

Can you show me what it means to say that:

"I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G."

for, say, {1}?

I just don't get it.

 

[WannabeNewton]

Can you visualize how open sets in $\mathbb{R}^{2}$ are all unions of open balls under the Euclidean topology? Can you picture an open ball in $\mathbb{R}^{2}$ under such a topology?

Don't feel bad about not being able to visualize topology in full generality! It's by no means easy and often times impossible. I find it usually helps to visualize things in euclidean 2-space. It will all come in due time :)

 

[micromass]

Yes, I looked back and had the discrete and indiscreet topologies mixed up.



Yes, my book has defined open balls. From what I can tell, it is just a set of y such that the "distance" between x and y is less than r.

The definition of open sets you gave was also given, but it's clear now that I should have spent more time on them, I'm having trouble getting a grasp over it.

It says that G is open if for any x in G, there is an r > 0 such that B(x,r) is a subset of G. So if for any x, there is an r such that the set of y such that the "distance" between x and y is less than r is a subset of G.

That is really confusing to me, I can't really make sense of it. I already understand open sets as a layman, in that it just "doesn't contain its boundary points" but I can't comprehend it in terms of open balls very well.

So, I can pick any element x of G,
and can find an r such that a set of y such that d(x,y) < r
is a subset of G.

Never before have I been able to type something and not "see it" or understand really what it implies at such a high degree. I really have no idea what this says about G.

The problem with the boundary interpretation is that you have not yet defined what a boundary is. You can define it precisely in function of open sets. So you can define the boundary using open sets, and then the boundary of an open set will not intersect the open set. But I guess this is not helpful, since it requires you to define open sets first.

My advice would be to find some examples of open sets in $\mathbb{R}$ and $\mathbb{R}^2$. And to verify explicitely that they're open. This would already help a lot.

The open ball interpretation says that if you take a point $x$ in the open set, then there is an entire collection of points near $x$ that is also in the open set. So in any "direction" you go from $x$, you will always encounter points in the open set and not other points (unless you go too far). This makes "not including the boundary" rigorous. Because if you were on the boundary, then you could go into a certain direction and not encounter any points in the set. Openness guarantees that there is no such point in the set. So no points in the set are in the boundary.

How would one do this?

Consider all subsets of X and then see it were true that "I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G." for each one?

Like, for the set X={1,2,3}
{{1}, {2}, {1,2},{2,3},∅, X}}

is a topology. So each element is an "open set."

Can you show me what it means to say that:

"I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G."

for, say, {1}?

I just don't get it.

The topology you give does not come from a metric space. So I cannot interpret it as function of open balls and metrics.

 

[1MileCrash]

Thanks for the advice you two, I am going to sleep on it and try some more of your suggestions and do some thinking on it tomorrow.

 

[WannabeNewton]

Sleep on it and compile a list of all the questions you have left! Have a good night.

 

[1MileCrash]

*sigh* I think I understand the ball definition of open sets, finally. All I had to actually draw a circle, call it a set, look at the definition, and I realize I was making it much more complicated than it needed to be.

Basically, it says that I can pick any element x in the set G, and I know that there are other elements in the set G that differ from x by a nonzero amount r. Even though the difference r is "radially," the important part is that it includes those elements that are nearer to the boundary of the set. So it's just saying that for any element x, there are elements that are closer to the boundary by a nonzero amount.

Is that right?

Even if I do understand that bit, it is still hard to imagine this making sense with finite sets.

 

[WannabeNewton]

What it's saying is that if you take any $x \in G$, where $G$ is an open set in a metric space, we can always find a sufficiently small radius $r$ such that all points within a distance $r$ from $x$ lie within $G$. This is exactly what you said if we take for granted that the notion of "boundary" has been defined. It's not as "geometric" with arbitrary topologies on finite sets, I agree, but picturing things in metric spaces can still help you a lot. By the way, all finite metric spaces are discrete so that should help you picture them better (try it).

 

[1MileCrash]

Ok, I'm still having a really hard time linking up the metric space and topological space.

They sound like two completely different ideas.

My book says that "every metric space is a topological space" but how does that make sense? A metric space is a set and a metric, which is a distance function. A topological space is a set and a collection of open subsets of that set.

If a metric space is also a topological space, then that would mean that this function d is a collection of subsets. How can you call a function with a numerical output a collection of subsets? How are these things equivalent?

 

[Fredrik]

My book says that "every metric space is a topological space" but how does that make sense?

Strictly speaking, that statement isn't true. But given a metric space (X,d), you can always define a topology on X by saying that a subset of X is open if and only if it's a it's union of open balls in X, with respect to d. If we define $\tau$ as the set of all subsets of X that are "open" in this sense, then $(X,\tau)$ is a topological space.

This is what people mean when they say that every metric space is a topological space.

 

[WannabeNewton]

If a metric space is also a topological space, then that would mean that this function d is a collection of subsets. How can you call a function with a numerical output a collection of subsets? How are these things equivalent?

As Fredrik said, the open balls of the metric form a topology generating basis i.e. the topology is that which is generated by the basis of open balls of the metric.

 

[1MileCrash]

Ok, I came up with this example for myself to try to understand the process, and to make my questions more clear.

Let X = { 1, 2, 3, 4, 5}
Let d be the normal metric d(x,y) = |x-y|
So (X, d) is a metric space.

I want to generate the induced topology. I know that the topology will be the collection of all open subsets of X.

How do I find all open subsets of X?

I listed every subset of {1,2,3,4,5} but I don't understand how to check if any particular subset is open.

Take for example, {1,2}

I know that this set is open if

"Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$."

There are two points in this subset to consider, 1 and 2.

B(1,r) has to be a subset of {1,2} for some r > 0.

B(1,r) = {y: |1-y| < r}

let r = .00000042

Then B(1,r) = {1}

and that's a subset of {1,2}

B(2,r) has to be a subset of {1,2} for some r>0

B(2,r) = {y: |2-y| < r}

let r = .000000042

Then B(1,r) = {2}

and that's a subset of {1,2}

So {1,2} is an open set.

What did I just say? It seems like I could make any set that I wanted open. Why can't I just always make r small enough so that B is always just the singleton set of x itself, which is of course a subset of the set I'm considering, because I picked it from the set? It's just not making any sense to me at all.

Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$. I assume these are all concepts you've already seen??

Anyway, take your metric space $(X,d)$. The topology generated by the metric space is simply the collection of all open sets.

This helps me understand what a topology is, and what an open set is, but I still don't see how I generate a topology from a metric. My only guess is that I have to list every single subset of X, and then look and see if there is an r such that the ball of radius r is a subset of X for every single element of every single subset of X, which according to my great lack of understanding above, there will be an r no matter what that satisfies that idea. This seems completely impossible to do anyway, my first exercise is to prove that the discrete metric generates the discrete topology, but I'm not told anything about the set its on, so what I've just described can't be the process. What is the process of generation?

 

[micromass]

Ok, I came up with this example for myself to try to understand the process, and to make my questions more clear.

Let X = { 1, 2, 3, 4, 5}
Let d be the normal metric d(x,y) = |x-y|
So (X, d) is a metric space.

I want to generate the induced topology. I know that the topology will be the collection of all open subsets of X.

How do I find all open subsets of X?

I listed every subset of {1,2,3,4,5} but I don't understand how to check if any particular subset is open.

Take for example, {1,2}

I know that this set is open if

"Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$."

There are two points in this subset to consider, 1 and 2.

B(1,r) has to be a subset of {1,2} for some r > 0.

B(1,r) = {y: |1-y| < r}

let r = .00000042

Then B(1,r) = {1}

and that's a subset of {1,2}

B(2,r) has to be a subset of {1,2} for some r>0

B(2,r) = {y: |2-y| < r}

let r = .000000042

Then B(1,r) = {2}

and that's a subset of {1,2}

So {1,2} is an open set.

What did I just say? It seems like I could make any set that I wanted open. Why can't I just always make r small enough so that B is always just the singleton set of x itself, which is of course a subset of the set I'm considering, because I picked it from the set? It's just not making any sense to me at all.

Right, any set is open here. So you could do the same with any set. Like already has been said, all finite metric spaces have the property that all sets are open. So if you want nontrivial examples, you need to look at infinite spaces.

This helps me understand what a topology is, and what an open set is, but I still don't see how I generate a topology from a metric.

Usually, you don't find the explicit topology at all, it's just too complicated. For example, you have no idea even what a generic open set in $\mathbb{R}^2$ looks like! However, you do know what the open balls look like. And the open balls generate the topology (in the sense that any open set is a union of open balls). So we say that the open balls form a basis. A lot of topological concept (actually, almost all) can be stated for bases instead of topologies. So the important concept in topology is actually a basis. A topology is usually too big.

My only guess is that I have to list every single subset of X, and then look and see if there is an r such that the ball of radius r is a subset of X for every single element of every single subset of X, which according to my great lack of understanding above, there will be an r no matter what that satisfies that idea. This seems completely impossible to do anyway, my first exercise is to prove that the discrete metric generates the discrete topology, but I'm not told anything about the set its on, so what I've just described can't be the process. What is the process of generation?

The set is arbitrary. So take an arbitrary set $X$ with the discrete metric. Now prove that any set is open.

Again, if you want more nontrivial examples, then look at open sets in $\mathbb{R}^2$.

 

[1MileCrash]

Thanks for your patience mm, so you mean that any subset of any finite set is always open regarding any metric on that set?

And, does this mean I am guaranteed to have a topology on a finite set if and only if I:

1.) Put the empty set and the entire finite set itself in the collection
2.) Take however many of whatever subsets I want, and put those in (the "basis?")
3.) Put the union/intersection of any number of any of the subsets I put in during (2), and put those in as well

But that's not true of infinite metric spaces because I couldn't just take whichever subset I wanted during (2), I'd have to verify it was open first, and then I could proceed in the same way?The next section is about topological basis. Do you think reading on would help me, or should I cover this better first?

The set is arbitrary. So take an arbitrary set $X$ with the discrete metric. Now prove that any set is open.

So it would be like,

Let (X,d) be the discrete metric space.

Let G be a subset of X, and g be an element of G.

G is open if there exists an r such that

B(g,r) = {y : d(g,y) < r } is a subset of GSo, I've tried proceeding here and I don't know what to do. Normally when I'm showing that something is a subset of something else, I pick an arbitrary element of the set and show it's in the other one.

Here I have to both show that any element of {y : d(g,y) < r } is a subset of G and show that there is an r that can make that happen. It's like I need one to prove the other, I don't see how to do these two things at once.

Is it the case that since d(g,y) is always 0 or 1, that I just need to show that any ball of radius > 1 is a subset of G? But wouldn't that mean that G would have to contain more than one element, thus not making G arbitrary?

Ugh, I'm so confused.

 

[micromass]

Thanks for your patience mm, so you mean that any subset of any finite set is always open regarding any metric on that set?

Yes.

And, does this mean I am guaranteed to have a topology on a finite set if and only if I:

1.) Put the empty set and the entire finite set itself in the collection
2.) Take however many of whatever subsets I want, and put those in (the "basis?")
3.) Put the union/intersection of any number of any of the subsets I put in during (2), and put those in as well

For metric spaces, you know that every set is open. So you need to end up with a topology that is every set. I don't see how your process guarantees that everything is open.

The next section is about topological basis. Do you think reading on would help me, or should I cover this better first?

I think you should verify some easy examples of topologies. Are there any exercises after this section? If not, I can provide you with some good exercises.

So it would be like,

Let (X,d) be the discrete metric space.

Let G be a subset of X, and g be an element of G.

G is open if there exists an r such that

B(g,r) = {y : d(g,y) < r } is a subset of G


So, I've tried proceeding here and I don't know what to do. Normally when I'm showing that something is a subset of something else, I pick an arbitrary element of the set and show it's in the other one.

Here I have to both show that any element of {y : d(g,y) < r } is a subset of G and show that there is an r that can make that happen. It's like I need one to prove the other, I don't see how to do these two things at once.

Is it the case that since d(g,y) is always 0 or 1, that I just need to show that any ball of radius > 1 is a subset of G? But wouldn't that mean that G would have to contain more than one element, thus not making G arbitrary?

So, you take $G$ arbitrary. You take $g\in G$ arbitrary. You need to find an $r$ such that $\{y~\vert~d(g,y)<r\}\subseteq G$. Focus on finding the right $r$ first.

Now, as a hint, for a given $r$, what are all the possibilities for $\{y~\vert~d(g,y)<r\}$. You will need to look at the cases $r\leq 1$ and $r>1$.

 

[1MileCrash]

Yes.



For metric spaces, you know that every set is open. So you need to end up with a topology that is every set. I don't see how your process guarantees that everything is open.

Sorry, I was a bit confused. I thought we had said that any subset of a finite metric space was automatically open, I need to reread.

I think you should verify some easy examples of topologies. Are there any exercises after this section? If not, I can provide you with some good exercises.

There are, I just don't think I'm in the position to attempt them. The last section, they started off really simple and got difficult, but I could do them all with time. The first one here, is a brick wall.

So, you take $G$ arbitrary. You take $g\in G$ arbitrary. You need to find an $r$ such that $\{y~\vert~d(g,y)<r\}\subseteq G$. Focus on finding the right $r$ first.

Now, as a hint, for a given $r$, what are all the possibilities for $\{y~\vert~d(g,y)<r\}$. You will need to look at the cases $r\leq 1$ and $r>1$.

Let r <= 1, then
d(g,y) < 1.

Since d is the discrete metric, that means that g = y and thus B(g,r) = {g}, which we know to be a subset of G since g is an element of G.

The other case is a bit confusing, I don't think what I say really makes sense.

--Ignore--

Let r > 1, then we have that

B(g,r) = {y : d(g,y) < r}

But if r is larger than 1, then the entire set of X is a subset of it, because the maximum distance allowed by the discrete metric is 1. Since X is a subset of B(g,r) if r > 1, then so is G.

WAIT, I need to show that B(g,r) is a subset of G, not the other way around. >< Ignore for now!

___

Can I say that the ball with r > 1 is "bigger" than X itself?

 

[micromass]

Sorry, I was a bit confused. I thought we had said that any subset of a finite metric space was automatically open, I need to reread.

Yes, any subspace of a finite metric space is open. So if $(X,d)$ is a finite metric space, then $\mathcal{T}= \mathcal{P}(X)$.

Let r <= 1, then
d(g,y) < 1.

Since d is the discrete metric, that means that g = y and thus B(g,r) = {g}, which we know to be a subset of G since g is an element of G.

Good. Doesn't this solve the exercise already?

The other case is a bit confusing, I don't think what I say really makes sense. If the radius of the ball is bigger than any d, can I say that the "boundary" of the ball lies outside of the set that the ball is in, so the set doesn't contain its boundary, so it is open?

Sorry, I have no idea what you're trying to say here.

Let r > 1, then we have that

B(g,r) = {y : d(g,y) < r}

But if r is larger than 1, then the entire set of X is a subset of it, because the maximum distance allowed by the discrete metric is 1. Since X is a subset of B(g,r) if r > 1, then so is G.

OK, so $X=B(g,r)$ in this case.

 

[1MileCrash]

Good. Doesn't this solve the exercise already?

Yeah, it does, because d is always <= 1.

So any ball for this metric is a singleton set for the center of the ball?

Sorry, I have no idea what you're trying to say here.

You can ignore it, I'm still forming thoughts and thinking out loud, it's a mess up there right now.

OK, so $X=B(g,r)$ in this case.

So clearly, we can't say that B(g,r) is a subset of G. But, then again, there is no such g,y such that d(g,y) > 1, so does it matter?

 

[micromass]

Yeah, it does, because d is always <= 1.

What do you mean? $d$ is a function, how can it be smaller than 1?

So any ball for this metric is a singleton set for the center of the ball?

Any ball with radius smaller or equal than $1$ is. If the radius is strictly larger than $1$, than the ball is the entire space.

So clearly, we can't say that B(g,r) is a subset of G. But, then again, there is no such g,r such that d(g,r) > 1, so does it matter?

What do you mean with $d(g,r)$?? I thought $r$ was a number. In that case $d(g,r)$ makes no sense.

 

[1MileCrash]

What do you mean? $d$ is a function, how can it be smaller than 1?

I mean that d(x,y) is less than or equal to 1 for any x,y. So why do we need to consider balls with a radius larger than the distance between any two points?

What do you mean with $d(g,r)$?? I thought $r$ was a number. In that case $d(g,r)$ makes no sense.

Sorry, that should be d(g,y).EDIT:

Darn, I just realized what you're saying. I need to affirm that an r EXISTS, and by showing that any r < 1 works for any arbitrary x in G, I've done just that. That's why the exercise is done.

 

[micromass]

I mean that d(x,y) is less than or equal to 1 for any x,y. So why do we need to consider balls with a radius larger than the distance between any two points?

We don't need to consider them for this problem. But it's still interesting to know what the balls are. So something like $B(g,2)$ exists here (and this set is equal to $X$). But it's not interesting. As you will soon find out, in topoogy we only care about balls with "small" $r$.