Show that $f:X\to Y$ is uniformly continuous $\iff f(N_{\delta,X}(x))\subset N_{\epsilon,Y}(f(x))$

Question

Suppose $X$ and $Y$ are metric spaces, and let $N_{\delta,X}(x)$ denote the $\delta$-neighborhood of $x$ with respect to the metric $d_X$ of the metric space $X$. Show that $f:X\to Y$ is uniformly continuous if, given $\epsilon>0,$ there exists $\delta>0$ such that $ f(N_{\delta,X}(x))\subset N_{\epsilon,Y}(f(x))$ for all $x\in X$.

This property seems logical, but I can't exactly wrap my head around why this is true. I've tried the following:

$$d_X(x_1,x_2)<\frac{1}{2}\delta \implies d_Y(f(x_1),f(x_2))<\frac{1}{2}\epsilon$$ which means that

$$x_1\in N_{\delta,X}(x_2) \implies f(x_1)\in N_{\epsilon,Y}(f(x_2))$$ as $f$ is uniformly contiuous. We also have, by the definition of functions, that:

$$x\in X \implies f(x)\in f(X)$$

Put this together, and you almost have that $f(x_1)\in f(N_{\delta,X}(x_2)) \implies f(x_1) \in N_{\epsilon,Y}(f(x_2))$. For this argument, I feel like you need the fact that $f$ is injective, so you can say that in fact:

$$x\in X \iff f(x)\in f(X)$$

which would give us $$f(x_1)\in f(N_{\delta,X}(x_2))\implies x_1\in N_{\delta,X}(x_2) \implies f(x_1) \in N_{\epsilon,Y}(f(x_2))$$

I feel like I'm thinking way out of the box to formulate this, while it seems like a simple identity rather than a theorem or something. How could you, properly, formulate this identity in a way that seems logical with respect to uniform continuity?

Answer 1

We don't need any $\frac{\varepsilon}{2}$ trick at all; the main point is the fact the same $\delta>0$ works at any point, and then you see that the condition is just a fancy restatement of definitions.

So suppose that

$$\forall \varepsilon>0:\exists \delta>0:\forall x \in X : f[N_{\delta,X}(x)] \subseteq N_{\varepsilon,Y}(f(x))\tag{1}$$

holds and we want to show uniform continuity:

Let $\varepsilon >0$ be given. Let $\delta > 0$ be as given by (1) for that $\varepsilon$.

Then if $x,x'$ are any points in $X$ with $d_X(x,x') < \delta$, then in particular $x' \in N_{\delta,X}(x)$ so $f(x') \in N_{\varepsilon,Y}(f(x))$ which means $d_Y(f(x'), f(x)) < \varepsilon$. So $f$ is uniformly continuous.

If $f$ is uniformly continuous, then (1) holds: let $\varepsilon > 0$ be given and let $\delta>0$ be chosen in accordance with the definition of uniform continuity. Then if $x' \in N_{\delta,X}(x)$ we know that $d_X(x,x') < \delta$ so that $d_Y(f(x), f(x')) < \varepsilon$ and so $f(x') \in N_{\varepsilon,Y}(f(x))$, and hence $f[N_{\delta,X}(x)] \subseteq N_{\varepsilon,Y}(f(x))$ as required.