A question involving sequential compactness and continuity of a function

[raw]

Homework Statement

Let f:M$$\rightarrow$$R be a function
I need to prove that if the graph of a function is compact then the function is continuous.

Homework Equations

We have defined compactness as follows: a set is compact if every sequence of a function has a subsequence which converges to a limit in the set.

The Attempt at a Solution

Let (p_n,y_n) be a convergent sequence in the graph of f where y_n = f(p_n) that converges to (p,y). I know that p_n converges to p in M and I'm assuming I need to use the sequential definition of continuity, that is if p_n=>p and f(p_n)=>f(p) then f is continuous. I'm also assuming that I have to use the face that a compact set is bounded. I just don't know how to use compactness to show that if p_n=>p and f(p_n)=>f(p).

 

[Dick]

You know (pn,f(pn)) has a convergent subsequence, say (pm,f(pm)) which converges to some point (p,M). The only issue is what if there is another convergent subsequence, say (pl,f(pl)) which converges to a different limit point, (p,L) where L is not equal to M. You know there is only one point on the graph with first element p, (p,f(p)). So?

 

[raw]

You know (pn,f(pn)) has a convergent subsequence, say (pm,f(pm)) which converges to some point (p,M). The only issue is what if there is another convergent subsequence, say (pl,f(pl)) which converges to a different limit point, (p,L) where L is not equal to M. You know there is only one point on the graph with first element p, (p,f(p)). So?

the graph is compact therefore it is closed so (pn,f(pn))converges to some point in the graph so all of its subsequences converge to the same point >>>(pl,f(pl)) would converge to the same point as (pm,f(pm)). I still don't see where to go from here

I also think that I need to use the fact that a compact set is closed AND bounded because the next part of the question asks me to find an example of a function whose graph is only closed but the function is discontinuous.

 

[micromass]

Try to use the following result

$$x_n\rightarrow x~\Leftrightarrow~\text{every subsequence}~(x_{k_n})_n~\text{contains a further subsequence which converges to}~x$$

It is not clear to me if your working with metric spaces or topological spaces. I'll assume the former. So I'll prove it for metric spaces:
Only the reverse implication requires a proof. So assume that xn does not converge to x. Then there exists an epsilon such that
$$\forall n:\exist m>n: |x_n-x|\geq \epsilon$$
We can now easily find a subsequence of x which does not come in B(x,epsilon) (the ball around x with radius epsilon). But the assumption says that a subsequence of this subsequence converges to x. Which can not be, since this subsequence also stays away from B(x,epsilon). Thus a contradiction is reached.


So all you need to prove now is that every subsequence of f(pn) contains a subsequence which converges to f(p).

 

[raw]

Try to use the following result

$$x_n\rightarrow x~\Leftrightarrow~\text{every subsequence}~(x_{k_n})_n~\text{contains a further subsequence which converges to}~x$$

It is not clear to me if your working with metric spaces or topological spaces. I'll assume the former. So I'll prove it for metric spaces:
Only the reverse implication requires a proof. So assume that xn does not converge to x. Then there exists an epsilon such that
$$\forall n:\exist m>n: |x_n-x|\geq \epsilon$$
We can now easily find a subsequence of x which does not come in B(x,epsilon) (the ball around x with radius epsilon). But the assumption says that a subsequence of this subsequence converges to x. Which can not be, since this subsequence also stays away from B(x,epsilon). Thus a contradiction is reached.


So all you need to prove now is that every subsequence of f(pn) contains a subsequence which converges to f(p).

Yes, I'm really having problems with the last part. How do I show that if f does not necessarily preserve limits?

 

[micromass]

Let $$p_n\rightarrow p$$. Take a subsequence $$f(p_{k_n})$$ of our original sequence. This relates to a sequence $$(p_{k_n},f(p_{k_n})$$. Since the graph is compact, there exists a further subsequence

$$(p_{k_{n_m}},f(p_{k_{n_m}})$$

which converges to a certain element (s,t). Since the graph is closed, we get that actually (s,t)=(s,f(s)). But since $$p_n\rightarrow p$$, we have that s=p. Thus our subsequence

$$(p_{k_{n_m}},f(p_{k_{n_m}})$$

converges to (p,f(p)).

By our theorem, we got that the entire sequence $$(p_n,f(p_n))$$ converges to (p,f(p)). Hence $$f(p_n)\rightarrow f(p)$$.

 

[Dick]

Yes, I'm really having problems with the last part. How do I show that if f does not necessarily preserve limits?

Did you follow what micromass was saying? If p_n->p and f(p_n) does NOT converge to f(p), then there is an e>0 and a subsequence of p_n, call it p_m, such that |f(p_m)-f(p)|>e. Are you ok with that? Since the graph is compact (p_m,f(p_m)) has a convergent subsequence. What do you conclude from that?

 

[raw]

Did you follow what micromass was saying? If p_n->p and f(p_n) does NOT converge to f(p), then there is an e>0 and a subsequence of p_n, call it p_m, such that |f(p_m)-f(p)|>e. Are you ok with that? Since the graph is compact (p_m,f(p_m)) has a convergent subsequence. What do you conclude from that?

OK I finally understand what he was trying to get me to do in the first post. Thank you both so much for the help!