Understanding Open/Closed Sets in $\mathbb{R}^n$

[evinda]

Hello! (Wave)

The following definition is given:

A set $U \subset \mathbb{R}^n$ is called open if for each $x \in U$ there is $B_d(x, \epsilon) := \{ y \in \mathbb{R}^n: d(x,y)< \epsilon\}$ -> open ball with center $x$ and radius $\epsilon$.

Could you explain me why the following set is open?

View attachment 4789

Why is the following set closed?

View attachment 4787Why is the following set neither open nor closed?

View attachment 4788

 

[I like Serena]

Hello! (Wave)

The following definition is given:

A set $U \subset \mathbb{R}^n$ is called open if for each $x \in U$ there is $B_d(x, \epsilon) := \{ y \in \mathbb{R}^n: d(x,y)< \epsilon\}$ -> open ball with center $x$ and radius $\epsilon$.

Could you explain me why the following set is open?

Why is the following set closed?

Why is the following set neither open nor closed?

Hey evinda! (Smile)

I think your definition should also include that the ball has to be a subset of $U$.

Anyway, it's all about the boundaries.

In your drawings it is not clear if the boundary is part of the set or not.
Presumably in your first drawing it's not, so that we have a ball around each of the inner points, such as $y$, that are indeed part of the set.

In the second drawing, I can only assume that the white parts represent open sets.
Therefore their complement, the shaded part, is closed.
For reference, a closed set is defined as the complement of an open set. (Nerd)

In the third drawing, any ball around (1,1) will not be a subset.
Therefore $D$ is not open.
The complement of $D$ has many points on the boundary for which no ball will be a subset.
It follows that the complement is not open, and thus $D$ is not closed. (Mmm)

 

[Deveno]

One way of characterizing "open" is in terms of "neighborhoods". This introduces another "layer" of definition, but it turns out to be easier to work with, in some instances.

So...what's a neighborhood? Well, a neighborhood is an element of a "system of neighborhoods". So...I know this may seem a bit repetitive...what's a neighborhood system?

Basically, a neighborhood system is the assignment of a filter $\mathcal{N}(x)$ (called, strangely enough, the neighborhood filter) to every element $x \in X$, where $X$ is the set we are studying subsets IN.

So...yes, wait for it...what's a filter?

Well, again, the typical way to proceed is to define a simpler structure called a filter base.

And...I hope you're still with me...we must ask: what is a filter base?

First, I interrupt this fascinating series of as-yet-undisclosed definitions, to make a comment:

The power-set of $X$, denoted $2^{X}$ or $\mathcal{P}(X)$ is a partially-ordered set, with the partial order (the anti-symmetric and reflexive operation "$\leq$") being given by inclusion: "$\subseteq$".

Ok, where was I?

A filter base on $X$ is a subset $B$ of $\mathcal{P}(X)$ such that:

1. $\emptyset \not\in B$
2. $B \neq \emptyset$
3. If $U,V \in B$ then there exists $W \in B$ with $W \subseteq U \cap V$ (this condition is called "downward-directed").

We turn this into a filter $F$ by adding the condition:

4. If $S \in F$ and $S \subseteq T$, then $T \in F$. (this condition is called "upwards closed").

Note this means that (combining 3 & 4) that $U,V \in F$ implies $U \cap V \in F$.

We can think of the neighborhood filter of a point $x$ as "sets that close in on $x$". Obviously, a *finer* filter, let's us get "closer".

What is our neighborhood filter base in this special case? The "$\epsilon$-balls" under the metric $d$ centered at $x$. In this case, then, a neighborhood $N(x)$ (completing the filter base to a filter) is a set $N(x)$ containing $x$ such that there exists an $\epsilon$-ball centered at $x$ with $B_d(x,\epsilon) \subseteq N(x)$.

Finally, we are ready to define "open":

A set $U$ is open if it is a neighborhood of each of its points.

To help determine which sets are open, and which are not, it is helpful to use two more concepts:

An interior point of a set $S$, is a point $x$ such that there exists a neighborhood $N(x)$ entirely within $S$. It should be clear that every point of an open set is an interior point, by definition. By the same token, if every point of $S$ is an interior point, $S$ is open.

A boundary point of a set $S$, is a point $x$ (not necessarily in $S$) such that EVERY neighborhood of $x$ intersects $S$ AND $X\setminus S$. The set of all boundary points of $S$ is denoted $\partial S$.

This gives the following useful result:

A set $S$ is open if and only if $S \cap \partial S = \emptyset$ (that is, $S$ does not include any of its boundary points).

This is easy to see: if $S$ does not include any boundary points, then if $x \in S$, there exists SOME neighborhood $N(x)$ of $x$ that does not intersect $S$ and $X\setminus S$. This neighborhood clearly intersects $S$ (since $x \in S$), and thus $N(x) \cap X\setminus S = \emptyset$, that is, $N(x) \subseteq S$, so every point is interior, and $S$ is open.

(we are using the basic result on sets: $A = (A\cap B)\cup (A\cap B^c)$, and $A\cap B = A \implies A \subseteq B$).

On the other hand, if $S$ is open, every point in $S$ is interior, and thus there exists a neighborhood of every point of $S$ contained entirely within $S$, and this neighborhood precludes any point of $S$ being a boundary point.

A closed set is one for which $S \cap \partial S = \partial S$, in other words, it contains all of its boundary points.

For example, the set $\{(x,y) \in \Bbb R^2: x^2 + y^2 < 1\} \cup \{(1,0)\}$ is neither open nor closed.

It is not open because $(1,0)$ is a boundary point, it is not closed because it does not contain the boundary point $(0,1)$ (along with many, many others).

(By the way, one cannot generally tell if a set is open or closed from a picture-boundaries are "pretty thin").

 

[evinda]

Hey evinda! (Smile)

I think your definition should also include that the ball has to be a subset of $U$.

Oh yes, I forgot to write it. I am sorry...

Anyway, it's all about the boundaries.

In your drawings it is not clear if the boundary is part of the set or not.
Presumably in your first drawing it's not, so that we have a ball around each of the inner points, such as $y$, that are indeed part of the set.

Could you maybe explain me how, given a graph of a set, we deduce if the given set is open or closed?
I haven't really understood how this can be done.. (Sweating)

 

[I like Serena]

Could you maybe explain me how, given a graph of a set, we deduce if the given set is open or closed?
I haven't really understood how this can be done.. (Sweating)

The standard convention is to use solid lines to indicate inclusion and dashed lines to indicate exclusion.
For instance like this:
View attachment 4790
View attachment 4791
(Happy)

Perhaps in your case a different convention was used.
If so that should be mentioned near the drawings, or in an introduction.

 

[evinda]

Perhaps in your case a different convention was used.
If so that should be mentioned near the drawings, or in an introduction.

In your drawings it is not clear if the boundary is part of the set or not.
Presumably in your first drawing it's not, so that we have a ball around each of the inner points, such as $y$, that are indeed part of the set.

So if we suppose that the boundary is not part of the set, then do we know that the set is open since the boundary is a ball with center $x$ and radius greater than $\epsilon$? Or have I understood it wrong? (Thinking)

 

[I like Serena]

So if we suppose that the boundary is not part of the set, then do we know that the set is open since the boundary is a ball with center $x$ and radius greater than $\epsilon$? Or have I understood it wrong? (Thinking)

The $x$ in the definition is different from the $x$ in the drawing. (Worried)

In the drawing we have some point $x$ with around it a set $U$ that may or may not be a ball.
Either way, it does not have radius $\epsilon$.

To apply the definition of an open set, we introduce a new symbol $y$ that takes the place of $x$ in the definition, since $x$ is already in use. (Nerd)
So for every $y$ in $U$, we need that there is some $\epsilon > 0$ however small, such that the mini ball around $y$ with radius $\epsilon$ fits inside $U$.
If the boundary of $U$ is not part of the set, this is always possible.

 

[evinda]

The $x$ in the definition is different from the $x$ in the drawing. (Worried)

In the drawing we have some point $x$ with around it a set $U$ that may or may not be a ball.
Either way, it does not have radius $\epsilon$.

To apply the definition of an open set, we introduce a new symbol $y$ that takes the place of $x$ in the definition, since $x$ is already in use. (Nerd)
So for every $y$ in $U$, we need that there is some $\epsilon > 0$ however small, such that the mini ball around $y$ with radius $\epsilon$ fits inside $U$.
If the boundary of $U$ is not part of the set, this is always possible.

I am a little confused right now. Do we suppose that only $x$ is given and $U$ is the area around $x$? Or have I understood it wrong? (Thinking)

 

[I like Serena]

I am a little confused right now. Do we suppose that only $x$ is given and $U$ is the area around $x$? Or have I understood it wrong? (Thinking)

Yes! (Smile)

 

[Deveno]

Oh yes, I forgot to write it. I am sorry...

Could you maybe explain me how, given a graph of a set, we deduce if the given set is open or closed?
I haven't really understood how this can be done.. (Sweating)

Naively, we look at the "edges". For a 1-manifold (a curve), this is the endpoints. For a 2-manifold (a surface) this is the perimeter. For a 3-manifold, this is the surface. In general, the boundary of an $n$-manifold is a $n-1$-manifold.

That said, this is the intuition of a very "special" kind of space (Euclidean space, with the Euclidean metric), and certain "simple" sets, built out of smooth maps of (respectively): intervals, disks, and spheres. In an arbitrary topological space, we only have the definitions to fall back on, as our intuition may fail us. Some very "bizarre" neighborhood systems can be constructed, with unexpected properties.

Not only that, but the power set of even the real line is...quite large. So there are a LOT of possible sets to examine. One cannot hope to be able to examine, or even classify, them all. Arbitrary functions of a real interval can be quite bizarre, and even limiting our attention to continuous functions isn't much better.

Open sets (in the real line these are unions of open intervals, in the real plane, these are unions of open disks...in the USUAL (metric) topology) is one way to "filter" this bewildering array of sets, to be able to speak about "homeomorphs" of regions (areas that are topologically equivalent, even though they may "look" radically different).

In analysis, metrics serve as a gauge of "good approximation". If a function stays within certain bounds, for example, we mean its graph (on the domain of definition) lies within a closed set. Ideally, we can show a sequence of functions lie within a sequence of ever-decreasing closed sets. Note "within" here means in the INTERIOR of said closed sets, and we can form a FILTER from the sequence of these interiors. If this filter is finer (i.e. contains more sets than) the neighborhood filter for our function space, we have a notion of CONVERGENCE. This allows us to take LIMITS of functions, not just limits of the individual points of definition, or individual image points. This is a VERY powerful tool.

Bound up with all of this, is the concept of "nearness" (spatial "closeness"). Here is a proper definition:

A point $P$ is near a set $A$ (written $A \leftarrow P$) if for every neighborhood $N(P)$, we have:

$N(P) \cap A \neq \emptyset$.

Some obvious consequences of this definition:

1. If $P \in A$, then $A \leftarrow P$.
2. If $P \in \partial A$, then $A \leftarrow P$.
3. If $P$ is in the interior of $X\setminus A$, then $P$ is not near $A$.

If every point of $A$ is not near $X\setminus A$, then $A$ is open. Near points can be thought of as the points in or "touching" (the corresponding geometric term is "tangent") a set $A$.

The re-casting of the term "continuous" in nearness-terms is particularly satisfying:

$f$ is continuous if $A \leftarrow P \implies f(A) \leftarrow f(P)$

capturing our usual intuition that continuous functions don't cause "tears" in the sets they transform.

Huh. It's a lot to process, eh?

It's really hard, as I said earlier, to just "look at the picture". To really "prove" a set is open, one often HAS to use the definitions. Here is an example:

Let's PROVE that the set $A = (0,1]$ is not open. We will consider two cases:

1. $x \in (0,1)$
2. $x = 1$

In case 1, we will show $\Bbb R\setminus A \leftarrow x$ is false.

We have: $0 < x < 1$, so let $\epsilon = \frac{1}{2}\min(x,1-x)$

Since $0<x$ and $x< 1$, so that $0 = 1-1 < 1-x$, we have $\epsilon > 0$ (half of a positive number is still positive).

Let $N(x) = (x-\epsilon,x+\epsilon) = B_d(x,\epsilon)$.

Then $N(x) \subseteq (0,1)$ but $N(x) \cap \Bbb R\setminus A = \emptyset$, so $x$ is not near $\Bbb R\setminus A$ (recall that if $x$ is near a set, EVERY neighborhood has to intersect it, and we found a counter-example).

So in case, 1, we have actually proven $(0,1)$ is open. But, we're not done yet.

Now suppose $x = 1$. Let $N(1)$ be ANY neighborhood of $1$. Then, since $N(1)$ is a neighborhood, there is some $B_d(1,\epsilon)$ for some $\epsilon > 0$, with $B_d(1,\epsilon) = (1-\epsilon,1+\epsilon) \subseteq N(1)$. Without loss of generality, if $\epsilon \geq 1$, we can replace it with $\frac{1}{\epsilon + 1}$ (this will still be inside $N(1)$), so we will assume that $\epsilon < 1$.

Now $1 - \epsilon < 1$, and $0 = 1 - 1 < 1 -\epsilon$, so for any $y \in (1-\epsilon,1) \subset B_d(1,\epsilon) \subseteq N(1)$ we have $y \in N(1)$ and $y \in A$, thus $N(1) \cap A \neq \emptyset$.

This shows $1$ is near $A$.

On the other hand, $1 + \epsilon > 1$, so for any $y \in (1,1+\epsilon) \subset B_d(1,\epsilon) \subseteq N(1)$, we also have $y \in \Bbb R\setminus A$, so $N(1) \cap \Bbb R\setminus A \neq \emptyset$.

This shows that $1$ is near the complement of $A$ in $\Bbb R$, that is, it is a boundary point of $A$ (near $A$ and its complement), and open sets do not contain any boundary points. Hence, $A$ is not open.

Now, this is a lot "harder" than just looking at a picture, but this is what is actually involved in "nailing it down air-tight".

 

[evinda]

Yes! (Smile)

So for every $y$ in $U$, we need that there is some $\epsilon > 0$ however small, such that the mini ball around $y$ with radius $\epsilon$ fits inside $U$.
If the boundary of $U$ is not part of the set, this is always possible.

So doesn't the fact whether there is a mini ball around $y$ with radius $\epsilon$ fits inside $U$ depend on $x$?

Also what happens if the boundary of $U$ is part of the set?

 

[Deveno]

So doesn't the fact whether there is a mini ball around $y$ with radius $\epsilon$ fits inside $U$ depend on $x$?

Also what happens if the boundary of $U$ is part of the set?

Yes, if $x$ is on the "very edge" of $U$ (the boundary), no ball around $x$ will "fit inside" $U$, it will "spill over" to the complement.

If $U$ contains part of its boundary (even just one point), it will NOT be open.

 

[evinda]

In the second drawing, I can only assume that the white parts represent open sets.
Therefore their complement, the shaded part, is closed.
For reference, a closed set is defined as the complement of an open set. (Nerd)

So do we assume that the boundary of the inner sets are not included?
If so, is the boundary included at the complement? (Thinking)

- - - Updated - - -

In the third drawing, any ball around (1,1) will not be a subset.
Therefore $D$ is not open.
The complement of $D$ has many points on the boundary for which no ball will be a subset.
It follows that the complement is not open, and thus $D$ is not closed. (Mmm)

Does it hold that any ball around $(1,1)$ will not be a subset because of the fact that $(1,1)$ is at the boundary?
In general, does it hold always if a point of the boundary is included in $D$ that $D$ will not be open?

Could you explain me further the following?

The complement of $D$ has many points on the boundary for which no ball will be a subset

- - - Updated - - -

Also there is the following definition:

A set $K \subset \mathbb{R}^n$ is called bounded if there is a $\Lambda>0$ such that for each $x \in K$,
$$d(x,0)< \Lambda \text{ or equivalently } K \subset B_d(0, \Lambda)$$

How can we apply this definition if we are given a graph of a set?

 

[I like Serena]

So do we assume that the boundary of the inner sets are not included?
If so, is the boundary included at the complement? (Thinking)

The boundary of the inner sets would be included in the shaded set.
That's what we need for it to be closed. (Nerd)

Does it hold that any ball around $(1,1)$ will not be a subset because of the fact that $(1,1)$ is at the boundary?
In general, does it hold always if a point of the boundary is included in $D$ that $D$ will not be open?

Yes. (Nod)

Could you explain me further the following?

The complement of $D$ has many points on the boundary for which no ball will be a subset

The complement of $D$ consists of everything outside the set, including the boundary, except for (1,1).
Since it includes points on the boundary, it cannot be open. (Thinking)

Also there is the following definition:

A set $K \subset \mathbb{R}^n$ is called bounded if there is a $\Lambda>0$ such that for each $x \in K$,
$$d(x,0)< \Lambda \text{ or equivalently } K \subset B_d(0, \Lambda)$$

How can we apply this definition if we are given a graph of a set?

If the graph fits on a piece of paper, obviously there is some $\Lambda>0$ that bounds it. (Mmm)