Theroem of closed sets containing limit

[Lee33]

Homework Statement

Theorem: Let S be a subset of the metric space E. Then S is closed iff whenever p1,p2,p3,... is a sequence of points of S that is convergent in E, we have lim n→∞ p_n ∈ S.

Homework Equations

The Attempt at a Solution

I am having trouble understand the "if" part of the proof. It goes:

Suppose S ⊂ E is not closed. Then S^c is not open and there exists a point p ∈ S^c such that any open ball of center p contains points of S. Hence for each positive integer n we can choose p_n ∈ S such that d(p,p_n) < 1/n. Then lim n→∞ p_n = p with p_n ∈ S and p not in S.

My question:

Why would there be an open ball of center p containing points of S? Also, why did the book choose 1/n for (what is d(p,p_n) < 1/n telling us)?

 

[gopher_p]

Homework Statement

Theorem: Let S be a subset of the metric space E. Then S is closed iff whenever p1,p2,p3,... is a sequence of points of S that is convergent in E, we have lim n→∞ p_n ∈ S.

Homework Equations

The Attempt at a Solution

I am having trouble understand the "if" part of the proof. It goes:

Suppose S ⊂ E is not closed. Then S^c is not open and there exists a point p ∈ S^c such that any open ball of center p contains points of S. Hence for each positive integer n we can choose p_n ∈ S such that d(p,p_n) < 1/n. Then lim n→∞ p_n = p with p_n ∈ S and p not in S.

My question:

Why would there be an open ball of center p containing points of S?

In a metric space, a set $U$ is open if and only if every for every element $x\in U$ there is an open ball centered at $x$ contained in $U$. This is for all intents and purposes by definition. The negation of this definition would say that a set $U$ is not open if and only if there is $x\in U$ such that every open ball centered at $x$ is not contained in $U$.

Also, why did the book choose 1/n for (what is d(p,p_n) < 1/n telling us)?

This is a common strategy for constructing sequences in metric spaces that converge to a specified element. In this case, we "know" that the ball $B(p,\frac{1}{n})$ contains at least one element of $S$ (it's actually many more than one), and so we chose one and call it $p_n$.

 

[Lee33]

Thanks for the feedback!

In a metric space, a set $U$ is open if and only if every for every element $x\in U$ there is an open ball centered at $x$ contained in $U$. This is for all intents and purposes by definition. The negation of this definition would say that a set $U$ is not open if and only if there is $x\in U$ such that every open ball centered at $x$ is not contained in $U$.

I am still a bit confused. The proof said that the point p ∈ S^c but then it says any open ball of center p contains points of S but how can points be in S^c and S? Since S^c is the complement then there can't be a point in both?

This is a common strategy for constructing sequences in metric spaces that converge to a specified element. In this case, we "know" that the ball $B(p,\frac{1}{n})$ contains at least one element of $S$ (it's actually many more than one), and so we chose one and call it $p_n$.

Why did they choose 1/n? Is it just because it is a convenient convergent sequence?

 

[gopher_p]

Thanks for the feedback!

No problem.

I am still a bit confused. The proof said that the point p ∈ S^c but then it says any open ball of center p contains points of S but how can points be in S^c and S? Since S^c is the complement then there can't be a point in both?

The assumption that $S^c$ is not open tells us that there is $p\in S^c$ such that every open ball centered at $p$ contains points not in $S^c$, i.e. in $S$.

Keep in mind that this is part of a proof by contradiction, so you need to forget for a moment everything that comes before "Suppose $S\subset E$ is not closed." The proof is designed so that what is actually going on does not match up with the consequences of that statement. It's possible that you are confusing yourself by trying to make sense of something that is supposed to be nonsensical. When you read a proof by contradiction, you kinda just need to play along until you get to the part that directly contradicts on of the assumptions of the claim; the part that says that $p$, which is not in $S$ is a limit of a sequence of points from $S$.

Why did they choose 1/n? Is it just because it is a convenient convergent sequence?

Basically, yes.

 

[Lee33]

The assumption that $S^c$ is not open tells us that there is $p\in S^c$ such that every open ball centered at $p$ contains points not in $S^c$, i.e. in $S$.

You think you can elaborate more on this part? This part is what I am having the most difficulty understanding.

 

[gopher_p]

You think you can elaborate more on this part? This part is what I am having the most difficulty understanding.

What is the definition of open set in a metric space?

 

[Lee33]

What is the definition of open set in a metric space?

The definition I use is, "a subset S of a metric space E is open if, for each p ∈ S, S contains some open ball of center p."

 

[gopher_p]

The definition I use is, "a subset S of a metric space E is open if, for each p ∈ S, S contains some open ball of center p."

Do you see that, given this definition, you can say ...

A subset $S$ of a metric space $E$ is not open if there is $p\in S$ such that $S$ does not contain an open ball centered at $p$

?

It has nothing to do with analysis or topology. It just the logical negation of your definition.

 

[Lee33]

Do you see that, given this definition, you can say ...

A subset $S$ of a metric space $E$ is not open if there is $p\in S$ such that $S$ does not contain an open ball centered at $p?$

It has nothing to do with analysis or topology. It just the logical negation of your definition.

I think I see it now. So it is safe to assume that if a set S is not open and it has a point $p\in S$ which contains an open ball then that open ball contains points not in $S$ hence in $S^c$?

If so, what kind of a point with an open ball will do that? The only point I can think of that will do that is a boundary point? Is that correct?

 

[gopher_p]

I think I see it now. So it is safe to assume that if a set S is not open and it has a point $p\in S$ which contains an open ball then that open ball contains points not in $S$ hence in $S^c$?

You need to be very precise with your language. It sounds like maybe you got it. But if I read this as though you mean every word of it literally it sounds like (a) you're maybe not aware that the empty set is open and (b) you think that points can contain open balls.

If so, what kind of a point with an open ball will do that? The only point I can think of that will do that is a boundary point? Is that correct?

You are correct. The point will be in the boundary. It is in the closure but not the interior.

 

[Lee33]

(b.) you think that points can contain open balls.

What type of points won't be able to contain an open ball? Just the empty set?

 

[gopher_p]

What type of points won't be able to contain an open ball? Just the empty set?

Open balls are special kinds of subsets of the metric space (which is a set with some additional structure). Points are elements of the metric space. Unless you're doing work in Set Theory, it typically does not make sense to say that an element of a set contains a subset of that set. It certainly does not make sense here to say that a point contains an open ball.

It does make sense to say that an open ball contains a point, though that is true for every open ball except the empty set (if it even makes sense to call the empty set an open ball; it is an open set). It would also make sense to say that a point is contained in a open ball, though that is true for every point in every metric space. In both cases it would probably still be better to say that the ball has an element or that the point is an element of the ball to avoid confusion with the notion of set containment.

I'm sorry if it seems like I'm being pedantic. It's just that so much of proving things and following the proofs of other boils down to understanding the meaning of the words that are being used.

 

[Lee33]

I see now, my use of the terms were not accurate. You've been a tremendous help. Thank you very much!