1. Unirormly Continuous Rectangles

Let $f:[a,b]$ x $[c,d]\rightarrow R$ be continuous on $[a,b]$ x $[c,d]$. Prove that $f$ is uniformly continuous on $[a,b]$ x $[c,d]$.

I understand the concept of uniform continuity. I'm assuming it can't be all that more complicated for a rectangle.

My idea is that if we can show that $f$ is uniformly continuous on [a,b] and $f$ is uniformly continuous on [c,d] then f is uniformly continuous on [a,b]x[c,d]???

Any help would be appreciated.

2. Originally Posted by zebra2147
Let $f:[a,b]$ x $[c,d]\rightarrow R$ be continuous on $[a,b]$ x $[c,d]$. Prove that $f$ is uniformly continuous on $[a,b]$ x $[c,d]$.

I understand the concept of uniform continuity. I'm assuming it can't be all that more complicated for a rectangle.

My idea is that if we can show that $f$ is uniformly continuous on [a,b] and $f$ is uniformly continuous on [c,d] then f is uniformly continuous on [a,b]x[c,d]???

Any help would be appreciated.
This really is the result of Heine-Cantor. Try using the Lebesgue Covering Lemma

EDIT: This is probably overkill, but I don't have time to think about it right now. There's absolutely an easier way that the one I mentioned above (maybe) if you're given that a continuous function on closed intervals is unif. cont. Let me think about it.

I prove the much more general result I alluded to earlier here (last theorem).

3. Well, I looked at all the links you gave me and they were all very interesting but I think I'm supposed to use the following lemma for this proof but extend it to rectangles...

Let $f:[0,1]\rightarrow \complex$ be continuous. Suppose $f$ is uniformly continuous on $I_{1}=[0,1/2]$, on $I_{2}=[1/4,3/4]$, and on $I_{3}=[1/2,1]$. Then $f$ is uniformly continuous on $[0,1]$.

That is why I previously stated my idea is to show that if $f$ is uniformly continuous on $[a,b]$ and is uniformly continuous on $[c,d]$ then f is uniformly continuous on $[a,b]$ x $[c,d]$??

4. Originally Posted by zebra2147
Well, I looked at all the links you gave me and they were all very interesting but I think I'm supposed to use the following lemma for this proof but extend it to rectangles...

Let $f:[0,1]\rightarrow \complex$ be continuous. Suppose $f$ is uniformly continuous on $I_{1}=[0,1/2]$, on $I_{2}=[1/4,3/4]$, and on $I_{3}=[1/2,1]$. Then $f$ is uniformly continuous on $[0,1]$.

That is why I previously stated my idea is to show that if $f$ is uniformly continuous on $[a,b]$ and is uniformly continuous on $[c,d]$ then f is uniformly continuous on $[a,b]$ x $[c,d]$??
Can you explain that lemma a little? It looks very strange and very arbitrary.

5. Well, I can try. He didn't cover it in class but the way I see it is that you have a continuous interval, I, and you break it into 3 seperate intervals $I_{1}, I_{2}, I_{3}$ and we say that all three of these intervals overlap so that every point in $I$ is contained in at least on of these subintervals. Then, if all three of these subintervals are uniformly continuous, then the interval $I$ must also be uniformly continuous.

6. Originally Posted by zebra2147
Well, I can try. He didn't cover it in class but the way I see it is that you have a continuous interval, I, and you break it into 3 seperate intervals $I_{1}, I_{2}, I_{3}$ and we say that all three of these intervals overlap so that every point in $I$ is contained in at least on of these subintervals. Then, if all three of these subintervals are uniformly continuous, then the interval $I$ must also be uniformly continuous.
And they, presumably, need to agree on the overlap, right?

7. Well, that isn't stated anywhere in his notes but based on what his lemma says, it would appear the overlap is consistent.

8. Originally Posted by zebra2147
Well, that isn't stated anywhere in his notes but based on what his lemma says, it would appear the overlap is consistent.
So are you allowed to extend this to arbitrary closed oovers. In other words, if $X$ is a space, $f:X\to\mathbb{R}$, and $\left\{X_{\alpha}\right\}_{\alpha\in\mathcal{A}}$ is a closed cover for $X$ such that $f_{\mid X_{\alpha}}:X_{\alpha}\to\mathbb{R}$ uniformly continuous then $f:X\to\mathbb{R}$ is uniformly continuous?

9. Well my professor never used the term "closed covers" but I looked up a little bit about it and that seems to be what is happening in his lemma. He also doesn't use discrete space terminology in his notes but what you wrote seems to work from what I gather.