Where i am wrong? A question on uniformly continuous function in functional analysis.

Given metric spaces $(X, d) $ and $(Y, d') $ , a mapping $T:X \to Y$ is uniformly continuous on $X$ iff for every Cauchy sequence $(x_n) $ in $(X, d) $, the sequence $ (Tx_n) $ is Cauchy in $(Y, d') $.

One sided implication (uniformly continuous function map Cauchy sequence to Cauchy sequence) is easy to prove.

But what about the converse.

I think it is false.

$f:\mathbb{R} \to \mathbb{R}$ defined by $$f(x) =x^2$$

maps Cauchy sequence to Cauchy sequence but this function is not uniformly continuous.

Please verify the proof that I upload as an image.

Is the proof correct?

Solution 1:

The prove is wrong (and cannot be repaired). If $T$ is not uniformly continuous, there is $r>0$ such that for all $n$ there are two points $x_n,y_n$ with $d(x_n,y_n)<1/n$ and $d'(Tx_n,Ty_n)>r$. One cannot generate a Cauchy sequence out of these points, as $(x_n,y_n)$ and $(x_{n+1},y_{n+1})$ are completely unrelated. This would only work if $(X,d)$ is compact