# Extending a uniformly continuous function

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To be honest, it isn't really all nonsense. Indeed, every normal space (in fact: every completely regular space) can be equipped with a notion of uniform continuity. But that is some more advanced topology.

So, what you wrote down is actually correct
Wow, that's interesting...

Yes, I'm sorry. That is what I meant. Stupid typo's of mine
No problem... I'm working on it..

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Hmm, the way you're approaching it is indeed to straightforward. Maybe you should try a proof by contradiction, i.e. assume that f is not uniformly continuous, then there exists sequences (x_n) and (y_n) such that d(x_n,y_n)->0 and d(f(x_n),g(x_n)) doesn't converge to 0.
Btw, another point - we said that uniform continuity of g is equivalent to dx(xn, yn) --> 0 ==> dy(g(xn), g(yn)) --> 0.

You mean, if we assume g not to be uniformly continuous, and if we'd find sequences such that dx(xn, yn) --> 0 ==> dy(g(xn), g(yn)) --> 0 holds, we'd arrive at a contradiction, right? (perhaps you made a type or I lack an understanding of logic...)

No, if g is not uniformly continuous, then there might still be sequences (x_n) and (y_n) such that d(x_n,y_n)-->0 and d(g(x_n),g(y_n))-->0.

What I'm claiming that (if g is not u.c.) there exists sequences (x_n) and (y_n) such that d(x_n,y_n)-->0 and such that d(g(x_n),g(y_n)) does not converge to 0.

The proof is actually not a proof by contradiction, but rather: a proof by contraposition. This is: a statement p => q is equivalent to (NOT q) => (NOT p).

So here: p = "for all sequences such that..." and q = "g is uniformly continuous"
So to prove p => q, it suffices to claim that "if q does not hold, then p does not hold".

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Ah, OK, now I understand it completely, thanks!

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No, sorry, but this doesn't seem to work for me.. :( Perhaps you can give me a hint...

So, g is uniform continuous if

$$\forall \epsilon >0: \exists \delta: \forall x,y: d(x,y)<\delta~\Rightarrow d(f(x),f(y))<\epsilon$$

So, g is not uniform continuous if there exists an $$\epsilon >0$$ such that

$$\forall \delta: \exists x,y: d(x,y)<\delta~\text{and}~d(f(x),f(y))\geq \epsilon$$

Now, take $$\delta=1/n$$, then

$$\exists x_n,y_n: d(x_n,y_n)<1/n~\text{and}~d(f(x),f(y))\geq \epsilon$$

This gives us the desired sequences (x_n) and (y_n) such that d(x_n,y_n)-->0 and d(f(x_n),f(y_n)) does not converge to 0.

Homework Helper
OK, if I got this right, for every ε we can find a δ and a pair of elements xn, yn such that dx(xn, yn) < δ implies dy(g(xn), g(yn)) >= ε. If we to this for every 1/n, where n is a positive integer, we arrive at a pair of sequences (xn) and (yn) whose distances converge to 0, but whose fodtances of images don't converge to 0, right?

Sorry, but sometimes I still have problems with formulating negations out of inequalities, I'm ashamed to admit.

OK, if I got this right, for every ε we can find a δ and a pair of elements xn, yn such that dx(xn, yn) < δ implies dy(g(xn), g(yn)) >= ε.
No, that is not correct. To negate a statement you need to change "forall" to "exists" and vice versa. Furthermore, the negation of "p=> q" is "p AND (NOT q)"

So, the negation is: "There exists en epsilon such that forall delta, there exists xn and yn such that dx(xn, yn) < δ and dy(g(xn), g(yn)) >= ε.

If we to this for every 1/n, where n is a positive integer, we arrive at a pair of sequences (xn) and (yn) whose distances converge to 0, but whose distances of images don't converge to 0, right?
This remains true.

Homework Helper
OK, I think I got it now. Thanks a lot. This was indeed an instructive and rigorous exercise!

Good! To be honest, this was not a very easy exercise...