Correctness of Statements about Set A in Metric Space X

[mathmari]

Hey! :giggle:

In $X=\mathbb{R}\setminus \mathbb{Q}$ with the usual metric, we consider $A=\{x\in X: \ 0\leq x\leq1\}$.
Which of the following statements are correct?
(a) $\partial A$ is singleton
(b) $A$ is open and closed
(c) $A$ is not open
(d) $A$ is not compact
(e) $A$ is as a metric space complete The boundary is $X\setminus A=\{x\in X: x<0 \lor x>1\}$, right? This is not a singleton, so (a) is not correct.
The boundary is not empty and so $A$ cannot be open and closed. So (b) is also not correct.
We have that $0\in \mathbb{Q}$ and $1\in \mathbb{Q}$, and so $A$ is not a closed interval, but an open one, or not? So (c) is also not correct.
A set is compact if it is closed and bounded. The set is not closed, and so it is not compact. So (d) is correct.
Is everything correct so far? How do we check (e) ?

:unsure:

 

[I like Serena]

In $X=\mathbb{R}\setminus \mathbb{Q}$ with the usual metric, we consider $A=\{x\in X: \ 0\leq x\leq1\}$.
Which of the following statements are correct?
(a) $\partial A$ is singleton

The boundary is $X\setminus A=\{x\in X: x<0 \lor x>1\}$, right? This is not a singleton, so (a) is not correct.

Hey mathmari!

For each statement we have to check the definition that applies.
Let's start with the boundary.
What you wrote is not the boundary $\partial A$, but the complement $A^c$. (Shake)

Check this wiki article for 3 common equivalent definitions of the boundary in topology.
If we pick one and apply it, can we deduce what $\partial A$ is?

 

[mathmari]

Check this wiki article for 3 common equivalent definitions of the boundary in topology.
If we pick one and apply it, can we deduce what $\partial A$ is?

Do we have to check if there are more than one irrational element in the set? :unsure:

 

[I like Serena]

Do we have to check if there are more than one irrational element in the set?

Huh?

Let's pick:

The boundary of a subset $A$ of a topological space $X$ is the set of points $p \in X$ such that every neighborhood of $p$ contains at least one point of $A$ and at least one point not of $A$: $\partial A := \{ p \in X \mid \forall O \ni p: O\cap A \neq \emptyset \,\text{ and }\, O \cap A^c \neq \emptyset \}$.​

Every $p$ in $A$ is irrational and has a small enough neighborhood such that all irrational points in it are in $A$.
Every neighborhood of both $0$ and $1$ have irrational points inside $A$ and also irrational points outside $A$. However, $0$ and $1$ are not in $X$, so they are excluded.
Every irrational point outside $[0, 1]$ has a small enough neighborhood such that all irrational points are outside $A$.

Therefore $\partial A=\varnothing$, which is not a singleton.

Since the boundary is empty, the argument for (b) that $A$ could not be open and closed does not hold.

Instead we might check whether $A$ is open and whether $X\setminus A$ is open.
A set in a topology is open if each of its points has a neighborhood that is contained in the set.

 

[mathmari]

Let's pick:

The boundary of a subset $A$ of a topological space $X$ is the set of points $p \in X$ such that every neighborhood of $p$ contains at least one point of $A$ and at least one point not of $A$: $\partial A := \{ p \in X \mid \forall O \ni p: O\cap A \neq \emptyset \,\text{ and }\, O \cap A^c \neq \emptyset \}$.​

Every $p$ in $A$ is irrational and has a small enough neighborhood such that all irrational points in it are in $A$.
Every neighborhood of both $0$ and $1$ have irrational points inside $A$ and also irrational points outside $A$. However, $0$ and $1$ are not in $X$, so they are excluded.
Every irrational point outside $[0, 1]$ has a small enough neighborhood such that all irrational points are outside $A$.

Therefore $\partial A=\varnothing$, which is not a singleton.

Ahh ok! :geek:

Since the boundary is empty, the argument for (b) that $A$ could not be open and closed does not hold.

A set is clopen if and only if its boundary is empty. That means that (b) holds, or not? :unsure:

Instead we might check whether $A$ is open and whether $X\setminus A$ is open.
A set in a topology is open if each of its points has a neighborhood that is contained in the set.

From the previous, that $A$ is clopen (open and closed), it follows that (c) cannot hold, can it? :unsure:
As for (d), a set is compact if it is closed and bounded. We have that $A$ is closed, so we have to check if it bounded, right? :unsure: How can we check (e) ? :unsure:

 

[I like Serena]

A set is clopen if and only if its boundary is empty. That means that (b) holds, or not?

Yep. (Nod)

From the previous, that $A$ is clopen (open and closed), it follows that (c) cannot hold, can it?

Correct. (Nod)

As for (d), a set is compact if it is closed and bounded. We have that $A$ is closed, so we have to check if it bounded, right?

Yes.

How can we check (e) ?

From wiki:

A metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M or, alternatively, if every Cauchy sequence in M converges in M.​

Is that the case? (Wondering)

 

[mathmari]

Yes.

A set is bounded iff it is contained inside some ball of finite radius $R$. In this case wecan consider the ballwith center the origin and radius $1$, right?
So we conslucde that $A$ is bounded.
Since $A$ is closed and bounded, it is compact.

Is that correct? :unsure:

From wiki:

A metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M or, alternatively, if every Cauchy sequence in M converges in M.​

Is that the case? (Wondering)

How can we check that? :unsure:

 

[I like Serena]

A set is bounded iff it is contained inside some ball of finite radius $R$. In this case we can consider the ball with center the origin and radius $1$, right?
So we conslucde that $A$ is bounded.
Since $A$ is closed and bounded, it is compact.

Is that correct?

Yep. (Nod)

How can we check that?

Can we find a Cauchy sequence of irrational numbers that has a rational number as its limit?

 

[mathmari]

Yep. (Nod)

:geek:

Can we find a Cauchy sequence of irrational numbers that has a rational number as its limit?

Can we use the sequence $a_n=2^{\frac{1}{n}+1}$ with limit $2\in \mathbb{Q}$ ? :unsure:

 

[I like Serena]

Can we use the sequence $a_n=2^{\frac{1}{n}+1}$ with limit $2\in \mathbb{Q}$ ?

I believe that works as a Cauchy sequence yes.
However, we also need that all elements of the sequence are inside $A$.
If we can achieve that, then we've proven that $A$ is not a complete metric space.

 

[mathmari]

I believe that works as a Cauchy sequence yes.
However, we also need that all elements of the sequence are inside $A$.
If we can achieve that, then we've proven that $A$ is not a complete metric space.

Ah you mean that for $n=1$ the element is not in $A$ ? :unsure:

 

[I like Serena]

Ah you mean that for $n=1$ the element is not in $A$ ?

Every element is greater than 1 so none of the elements are in $A$.

 

[mathmari]

Every element is greater than 1 so none of the elements are in $A$.

I haven't really understood that. Could you explain that further to me? Which elements are not in $A$ ? :unsure:

 

[I like Serena]

I haven't really understood that. Could you explain that further to me? Which elements are not in $A$ ?

For $x>0$ we have that $2^x>1$ don't we? So $2^{\frac 1n+1}$ is not in $A$.
Instead we can use $2^{\frac 1n-1}$ for $n\ge 2$. :geek:
We need to exclude $n=1$ because $1$ is not in $A$, and $1$ is also rational.

 

[mathmari]

For $x>0$ we have that $2^x>1$ don't we? So $2^{\frac 1n+1}$ is not in $A$.
Instead we can use $2^{\frac 1n-1}$ for $n\ge 2$. :geek:
We need to exclude $n=1$ because $1$ is not in $A$, and $1$ is also rational.

Ah yes!

So to show that this sequence is Cauchy we have to show that the difference between two terms is very small, right?
We have $$a_n-a_m=2^{\frac 1n-1}-2^{\frac 1m-1}=\frac{2^{\frac 1n}}{2}-\frac{2^{\frac 1m}}{2}=\frac{2^{\frac 1n}-2^{\frac 1m}}{2}$$ How do we continue?
We have that this sequence converges to $2^{-1}=\frac{1}{2}\notin A$, right?

:unsure:

 

[I like Serena]

Ah yes!

So to show that this sequence is Cauchy we have to show that the difference between two terms is very small, right?
We have $$a_n-a_m=2^{\frac 1n-1}-2^{\frac 1m-1}=\frac{2^{\frac 1n}}{2}-\frac{2^{\frac 1m}}{2}=\frac{2^{\frac 1n}-2^{\frac 1m}}{2}$$ How do we continue?
We have that this sequence converges to $2^{-1}=\frac{1}{2}\notin A$, right?

I believe it suffices that the limit exists, albeit not in $X$.
It means the elements of the sequence get arbitrarily close to each other, and since they are all in $A$, they fprm a Cauchy sequence in $A$.

 

[mathmari]

Let's pick:

The boundary of a subset $A$ of a topological space $X$ is the set of points $p \in X$ such that every neighborhood of $p$ contains at least one point of $A$ and at least one point not of $A$: $\partial A := \{ p \in X \mid \forall O \ni p: O\cap A \neq \emptyset \,\text{ and }\, O \cap A^c \neq \emptyset \}$.​

Every $p$ in $A$ is irrational and has a small enough neighborhood such that all irrational points in it are in $A$.
Every neighborhood of both $0$ and $1$ have irrational points inside $A$ and also irrational points outside $A$. However, $0$ and $1$ are not in $X$, so they are excluded.
Every irrational point outside $[0, 1]$ has a small enough neighborhood such that all irrational points are outside $A$.

Therefore $\partial A=\varnothing$, which is not a singleton.

I read again this part and I got stuck.

Could you eplain to me why every irrational point outside $[0, 1]$ has a small enough neighborhood such that all irrational points are outside $A$ ? Does it mean for example if we consider the irrational $\sqrt{2}$ and we take a neighborhood then all irrational of that neighborhood are outside $A$ ? :unsure:

And how does it follow that $\partial A=\varnothing$ ? :unsure:

 

[I like Serena]

I read again this part and I got stuck.

Could you eplain to me why every irrational point outside $[0, 1]$ has a small enough neighborhood such that all irrational points are outside $A$ ? Does it mean for example if we consider the irrational $\sqrt{2}$ and we take a neighborhood then all irrational of that neighborhood are outside $A$ ?

Suppose we pick the neighborhood $S=\{x\in X: \sqrt 2-0.1<x<\sqrt 2 +0.1\}$.
Then all points in $S$ are outside $A$.
Therefore $\sqrt 2$ is not in the boundary of $A$, because every neighborhood must have both a point inside and a point outside of $A$.

And how does it follow that $\partial A=\varnothing$ ?

The only points for which all neighborhoods in $X$ have points inside and outside of $A$, are $0$ and $1$.
However, those points are not in $X$, so they are not part of the boundary after all.

 

[mathmari]

Suppose we pick the neighborhood $S=\{x\in X: \sqrt 2-0.1<x<\sqrt 2 +0.1\}$.
Then all points in $S$ are outside $A$.
Therefore $\sqrt 2$ is not in the boundary of $A$, because every neighborhood must have both a point inside and a point outside of $A$.

Ah ok! And to consider the other case, picking a point in $A$ with a neighborhood inside $A$ can we consider $p=\frac{\sqrt{2}}{2}$ ? Then $S=\{x\in X: \frac{\sqrt{2}}{2}-0.1<x<\frac{\sqrt{2}}{2} +0.1\}$ and all point in $S$ are inside $A$.

Is that correct? :unsure:

 

[I like Serena]

Ah ok! And to consider the other case, picking a point in $A$ with a neighborhood inside $A$ can we consider $p=\frac{\sqrt{2}}{2}$ ? Then $S=\{x\in X: \frac{\sqrt{2}}{2}-0.1<x<\frac{\sqrt{2}}{2} +0.1\}$ and all point in $S$ are inside $A$.

Is that correct?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great! So is the only correct one the statement (b) ? :unsure:

It was a multiple choice online exercise... Ichose the statement (b) as the correct answer and I got the result that this is partially correct, that there is also more than one correct statements. But we have proven that allothers than (b) are wrong, or not? :unsure:

 

[I like Serena]

Great! So is the only correct one the statement (b) ?

I'm looking again and I believe (c) is also correct.
That is, $A$ is closed and bounded and therefore compact.

 

[mathmari]

I'm looking again and I believe (c) is also correct.
That is, $A$ is closed and bounded and therefore compact.

Statement (c) is that $A$ is not open. But at (b) we have that $A$ is open, how can that be? Or is $A$ not open because $A$ is open and closed? :unsure:

 

[I like Serena]

Statement (c) is that $A$ is not open. But at (b) we have that $A$ is open, how can that be? Or is $A$ not open because $A$ is open and closed? :unsure:

Oh wait. (Wait)

I misread. I meant that $A$ is compact, but (c) says that $A$ is not compact.
Therefore (c) is indeed false.

Either way, $A$ is both open and closed. For $A$ to be compact, we have the condition that it must be closed. And it is.

 

[mathmari]

Oh wait. (Wait)

I misread. I meant that $A$ is compact, but (c) says that $A$ is not compact.
Therefore (c) is indeed false.

Either way, $A$ is both open and closed. For $A$ to be compact, we have the condition that it must be closed. And it is.

So (c) is false.

Since $\partial A$ is the empty set, it is not a singleton, so (a) is also false.

What about (c) ? "$A$ is not open". When $A$ is open and closed, is it maybe true that it is not open? :unsure:

Since $A$ is compact statement (d) is also false.

For statement (e) we found a couterexample, so it is also false.

:unsure:

 

[I like Serena]

So (c) is false.

Since $\partial A$ is the empty set, it is not a singleton, so (a) is also false.

What about (c) ? "$A$ is not open". When $A$ is open and closed, is it maybe true that it is not open?

Since $A$ is compact statement (d) is also false.

For statement (e) we found a couterexample, so it is also false.

I mixed up (c) and (d). (Blush)

$A$ is both open and closed.
So saying that $A$ is not open is false.

 

[mathmari]

I mixed up (c) and (d). (Blush)

$A$ is both open and closed.
So saying that $A$ is not open is false.

Ok! That means only (b) can be correct, right? :unsure:

 

[I like Serena]

Ok! That means only (b) can be correct, right?

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Thank you for your help!

 

[Euge]

I do not see how $A$ has empty boundary. The closure of $A$ is $[0,1]$ since the irrationals are dense in the reals, and $A$ has empty interior since the rationals are dense in the reals. So wouldn't $\partial A$, being $\bar{A} - \operatorname{Int}(A)$, equal $[0,1]$?

 

[Euge]

Furthermore, since $A$ has empty interior, $A$ cannot be open, making (c) the correct answer.

 

[I like Serena]

I do not see how $A$ has empty boundary. The closure of $A$ is $[0,1]$ since the irrationals are dense in the reals, and $A$ has empty interior since the rationals are dense in the reals. So wouldn't $\partial A$, being $\bar{A} - \operatorname{Int}(A)$, equal $[0,1]$?

The topology is $X=\mathbb R\setminus\mathbb Q$ with the usual metric.
So none of the rationals can be part of either the interior or the boundary.
So the closure of $A$ is $[0,1]\setminus \mathbb Q$, which is the same as the interior of $A$, making the boundary $\partial A$ empty, doesn't it?

 

[Euge]

Looking back at the original question, it appears that I overlooked that $A$ inherits the subspace topology relative to $X$, not to $\Bbb R$. My apologies. Certainly $A = (0,1) \cap X$, which is relatively open in $X$, and $X - A = (\Bbb R - \Bbb [0,1]) \cap X$, which is relatively open in $X$; thus, $A$ is both open and closed in $X$, so that $A$ has empty boundary.