Extending a uniformly continuous function

[micromass]

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.

 

[radou]

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.

 

[micromass]

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.

 

[radou]

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

 

[micromass]

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