Let $f:X \rightarrow Y$ uniformly continuous. if $A$ is bounded in $X$ the $f(A)$ is bounded in $Y$

Question

Let $f:X \rightarrow Y$ uniformly continuous. if $A$ is bounded in $X$ the $f(A)$ is bounded in $Y$

I said it is true because:

Let A is bounded in X then exists $ k> 0 $ such that $d_{X}(x,y)\leq k$ for all $x,y \in A$. Let $\epsilon>0 $ then exists $\delta_{1}>0$ such that $d_{X}(a,b)<\delta_{1}$ then $d_{Y}(f(a),f(b))<\epsilon$ for all $a,b \in X$, so take $\delta=min\{k, \delta_{1}\}$. Now, let $v,w \in f(A)$ then exists $x_{1}$, $x_{2}$ $\in A \subseteq X$ such that $f(x_{1})=v$ and $f(x_2)=w$; then for $d_{X}(x_1,x_2)<\delta$ then $d_{Y}(v,w)=d_{Y}(f(x_1),f(x_2))<\epsilon$ for all $u,v \in f(A)$. So, $f(A)$ is bounded in $Y$

I want to confirm if it's correct what I did.

Answer 1

As mentioned by K.Power, this does hold whenever $X$ has the Heine-Borel property.

Here's an explicit counterexample:

Consider a new metric defined on $\mathbb{R}$ by $$ d(x,y):=\frac{|x-y|}{1+|x-y|} $$

I leave the verification that this actually is a metric as an exercise to you (or Google).

The map $t\mapsto \frac{t}{1+t}$, going from $[0,\infty)$ to $[0,1)$, sends $0$ to $0$ and has a continuous inverse (continuity of the inverse at $0$ is sufficient). Using this, one can see that:

A function $f:\mathbb{R}\to\mathbb{R}$ is uniformly continuous (when viewed with the standard Euclidean metric) if and only if it is uniformly continuous when viewed as a function $f:(\mathbb{R},d)\to \mathbb{R}$, where we consider the Euclidean metric on the codomain.

Armed with this knowledge, consider the identity function $$ \text{id}:\mathbb{R}\to\mathbb{R} $$ i.e. $\text{id}(x)=x$. This is obviously uniformly continuous, and therefore will also be uniformly continuous when viewed as a map $$ \text{id}:(\mathbb{R},d)\to \mathbb{R} $$ However, the entire space $(\mathbb{R},d)$ is bounded, simply because $d(x,y)<1$ for all $x,y$, whereas $\text{id}[\mathbb{R}]=\mathbb{R}$ is unbounded with the Euclidean metric.

Answer 2

This proof is not correct. Your $x_1$ and $x_2$ depend on your $u,w\in f(A)$. As such you are not guaranteed that $d_X(x_1,x_2)<\delta$. You have fixed an arbitrary $\varepsilon>0$, and as such it is unlikely (generally impossible) that you'll be able to prove that $d_Y(u,w)<\varepsilon$ for all $u,w\in f(A)$.

In fact this is not true for arbitrary metric spaces, but it is true for $\mathbb R$, as this post shows. For a counterexample to the general statement I would consider a metric space not satisfying the Heine-Borel property, such a space is described here.