# Simple proof of continuity of a metric space

1. Oct 23, 2011

### chels124

1. The problem statement, all variables and given/known data

Let X and Y be metric spaces, f a function from X to Y:
a) If X is a union of open sets Ui on each of which f is continuous prove that f is continuous on X.
b) If X is a finite union of closed sets F1, F2, ... , Fn on each of which f is continuous, prove that f is continuous on X.
c) Can part (b) be extended to an infinite number of closed subsets?

2. Relevant equations

The following three are equivalent:
(i) f is continuous
(ii) The complete inverse image of an open set is open
(iii) The complete inverse image of a close set is closed.

3. The attempt at a solution

(summarized obviously)
A) Is it enough to say that we know for each open set that makes up X, there exists an inverse image of that set in Y. Since this is equivalent to continuity, we know f is continuous on X.
B) Similarly, we know for each closed set that makes up X, there exists an inverse image of that set in Y. Since this is equivalent to continuity, we know f is continuous on X.
C) This is the part that confused me and made me feel like (A) and (B) were incorrect. I saw somewhere that it cannot be extended to an infinite number of closed subsets, but I don't understand why. If we use the equivalences from above, why wouldn't the function be continuous? So, I thought maybe I summed everything up a bit too quickly, but I can't find my own contingencies.

Thanks for any help!

2. Oct 23, 2011

### canis89

I'll try to help for 1a) first.

By your assumption, f maps X to Y. So, an inverse image of f must be a subset of X, not Y(or maybe you did a typo there?).

Hint 1. Try to relate the statement f is continuous on Ui with the statement (ii):

Hint 2. Since X is composed of Ui's, the range of f can be expressed as

$R(f)=\bigcup_i f(U_i)$.

Try to find a sequence of mutually exclusive sets Ai such that

$\bigcup_i f(U_i)=\bigcup_i A_i$, and $A_i\subseteq f(U_i)$.

After that, use Hint 1. Try to continue from there.

Edit: Sorry, I think this strategy only works for countable union of Ui's. Let me think of something else.

Last edited: Oct 24, 2011
3. Oct 24, 2011

### HallsofIvy

Staff Emeritus
b) requires finite union because an infinite union of closed sets is not necessarily closed.