Continuous functions on metric spaces with restrictions

[tjackson3]

Homework Statement

Let E,E' be metric spaces, f:E\rightarrow E' a function, and suppose that S_1,S_2 are closed subsets of E such that E = S_1 \cup S_2. Show that if the restrictions of f to S_1,S_2 are continuous, then f is continuous. Also, if the restriction that S_1,S_2 are closed is dropped, does the assertion still hold? Explain.

Homework Equations

Definition of continuity of a function from (E,d) to (E',d') at a point p_0: A function is continuous at p_0 if for any \epsilon > 0 and p \in E, there exists \delta > 0 such that if d(p,p_0) < \delta, d'(f(p),f(p_0)) < \epsilon.

The Attempt at a Solution

Let p, p_0 \in E and let \epsilon > 0. Then there are two cases to consider:

1.) Since in S_1,S_2, for any \epsilon > 0, there exists \delta > 0 such that if p \in S_1 (or equivalently S_2), if d(p,p_0) < \delta, d'(f(p),f(p_0)) < \epsilon by the definition of continuity. If, for the given choice of \epsilon, the corresponding \delta is such that B_{\delta}^E(p_0) = B_{\delta}^{S_1}(p_0) (i.e. if the ball of radius \delta is entirely contained in S_1), then by hypothesis, f is continuous at p_0. The same logic applies for S_2.

2.) Now suppose that the corresponding \delta is such that B_{\delta}^E(p_0) \neq B_{\delta}^{S_1}(p_0). In other words, some of the elements of the ball computed in E are in S_1, while some are in S_2.

I'm not sure where to go from here. I have an idea (just separating into two separate balls those elements of S_1,S_2 and applying the hypothesis), but this idea does not use the fact that the two sets are closed.

As for the final part of the question, if both sets are open, then f is continuous (due to a theorem involving the preimage of a function being open implying that f is continuous), but I feel like if one of the sets is open, while the other is closed, the assertion wouldn't hold. I think figuring out how to involve the fact that the sets are closed into the main part of the proof would go a long way to figuring this out, but I'm not sure.

Thanks so much for your help!

 

[micromass]

Have you seen that a function iss continuous iff every inverse image of a closed/open set is closed/open? That would make it a lot easier...

 

[tjackson3]

I've seen that that is the case iff the inverse image of an open set is open. To prove it for closed sets, is it just a matter of replacing the word "open" with "closed" in the proof for the open case, or is it more complicated than that?

 

[micromass]

No, I don't think replacing the word open with closed would suffice. But it's not that hard to show:

F~\text{closed}~\Rightarrow~Y\setminus F~\text{open}~\Rightarrow~f^{-1}(Y\setminus F)=X\setminus f^{-1}(F)~\text{open}~\Rightarrow~f^{-1}(F)~\text{closed}

The other implication is analogous.

So a map is continuous if the inverse image of a closed set is closed. Can you prove it now?

 

[tjackson3]

Easy enough. Thank you so so much! I'm still not sure how to show that if one of those subsets is closed and the other is open, the function is not continuous, though. Any ideas on that?

 

[micromass]

Can you give an arbitrary discontinuous function. It will probably be piecewise defined. Can you split that domain up in an open and a closed set?

 

[tjackson3]

Gotcha. Thanks again!