Proving Closed Sets using the Sequential Criterion

[Bleys]

I'm sorry if this should be in the Analysis forum; I figured it pertained to topology though.
Let Y be a subspace of a metric space (X,d) and let A be a subset of Y. The proposition includes conditions for A to be open or closed in Y. In class the teacher first proved when A is open and then used complements to prove when A is closed. I'm trying to go the other way around and prove the case when A is closed first, using the sequence criterion. The statement is:
A is closed in Y iff there is a closed set C in X such that $$A=C \cap Y$$
I haven't really used the sequential criterion for closed sets before, so I don't really know where to start :/ can you provide a starting point? Do I need to construct C or can I abstractly prove the existence of one?

 

[quasar987]

I'm sure you can at least unwind the definitions and write what it is you need to prove in each of the directions "==>" and "<==".

This is always a good place to start in a proof.

 

[Bleys]

Well, for <=
Suppose $$A=C \cap Y$$ for C closed in X. Then every convergent sequence in C has its limit in C. Let $$x_{n}$$ be a convergent sequence in $$C \cap Y$$. In particular $$x_{n}$$ lies in C and hence so does its limit. Then it follows (does it?) $$lim_{n \rightarrow \infty} x_{n} \in C \cap Y$$. Hence A is closed in Y.

For => I'm still not sure, even with the definition.
Suppose A is closed in Y. Then every convergent sequence in A lies in A. Should I use a contradiction argument and use the fact Y\A is open?

 

[quasar987]

You are forgetting an important detail in the definitions.

A set Y be a subset of a metric space (X,d) and A a subset of Y. Then A is closed in Y if, by definition, for all sequences $(y_n)$ with $y_n\in A$ for every n that converges in Y, then that limit is in A.

What does in mean that a sequence $(y_n)$ converges in Y? It means that there exists y in Y such that $\lim_nd(y_n,y)=0$.

For exemple, take X=R with the usual metric d(x,y)=|x-y|, Y=(0,2), A=(0,1]. Then A is not closed in X since the sequence 1/n converges in X to 0, but 0 is not in A. But, as is easily verified, A is closed in Y. The above argument used above to prove that A is not closed in X does not apply here because 0 is not in Y, so according to the definition above, 1/n does not converge to 0, in Y (!)

So, with that said, a better formulation of the problem for the direction <== would be:
"Suppose A=C n Y for C closed in X. Then every sequence in C that converge in X has its limit in C. Let $(x_n)$ be a sequence of elements of A that converge in Y, to say, y. We must show that y belongs to A."

Try solving this, and also rewrite what the hypotheses is and what you want to show for the direction "==>".

 

[WWGD]

As a bit of (probably useless) trivia, given a topological space X, and a subspace A

you can identify limit points (more precisely, points in CL(A); the closure of A) with

sequences, iff X is first-countable , i.e., if it has a countable local basis. This, of

course, includes metric spaces. Specifically, in X 1st-countable, we can say that

a is in CL(A) iff there exists a sequence in A , converging to a. Outside of first-

countable, we can generalize with nets.

 

[Bleys]

$$x_{n} \in A$$ implies $$x_{n}$$ is in C and Y. Now the limit of the sequence is in Y, so in order to show y is in A, we have to show y is in C. But in particular $$x_{n}$$ is in X and since C is closed its limit will lie in C, so y is in C, hence is in A, as required.

Now, for ==> suppose A is closed in Y. That is, if $$x_{n} \in A$$ for all n is convergent in Y, then its limit is in A. Let C be the set of all sequences in A that converge in X, and its limits. By definition C is closed in X. Then $$C \cap Y$$ are all sequences in A that converge in Y, and its limit points. But this is just A. Is this correct?

 

[quasar987]

Gorgeous!

 

[Bleys]

Awesome, thanks for the help quasar!