Let $X$ and $Y$ be metric spaces and $f:X\longrightarrow Y $

Question

Proof that $f$ is uniformly continuous function if and only if for any sequence $(x_n)$ and $(y_n)$ in $X$ that $d_X(x_n,y_n)\rightarrow0$, we have $d_Y(f(x_n),f(y_n))\rightarrow0$

$"\Longrightarrow"$

Let's suppose we know $f:X\rightarrow Y $ and $f$ is uniformly continuous function.

Let's $(x_n)$ and $(y_n)$ in $X$ and $$d_X(x_n,y_n)\rightarrow0$$

It is correct to affirm that for $n\in \mathbb{N}$? $$x_n\longrightarrow y_n $$ If we can do that , applying $f$ because is uniformly continuous i.e $f$ is continuous :$$f(x_n)\longrightarrow f(y_n)$$ And we can see this is equivalent to: $$d_Y(f(x_n),f(y_n))\rightarrow0$$ $"\Longleftarrow"$

I think that by contradiction could be done.

Help me please.

Answer 1

The forward direction is complete nonsense: $x_n \to y_n$ is not defined.

Just take $\varepsilon > 0$. We need to find $N$ such that

$$\forall n \ge N: d_Y(f(x_n), f(y_n)) < \varepsilon$$

Find $\delta>0$ for uniform continuity with this $\varepsilon$. How do you get the $N$ from your unused assumption ?

For the reverse, use the contrapositive. Write out what it means to be not uniformly continuous, in logical terms. The sequences $(x_n)$ and $(y_n)$ to choose will hopefully suggest themselves, as soon as your realise that you can take epsilons and delta's of the form $\frac{1}{n}$ in that negation..

Answer 2

For a given positive number $\varepsilon >0$, we seek a positive real $\delta >0$ so that $d_Y (f(x),f(y))<\varepsilon $ whenever $d_X (x,y)<\delta $. Assume we cannot do this, then there is $\varepsilon_0 >0$ so that for each $n\in\mathbb{N}_{>0} $ and an element $x_n\in X$ there is an element $y_n\in X$ such that $d_X (x_n ,y_n )<\frac{1}{n} $ but $d_Y (f(x_n ),f(y_n ))\geq\varepsilon_0 $. This is absurd.