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

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?

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.

• The last part I don't quite understand, still. From what I can understand, in the last part you proved that $x'\in N_{\delta ,X}(x) \implies f(x')\in N_{\epsilon,Y}(f(x))$, while I don't see where you have $f(x')\in f[N_{\delta ,X}(x)]$ in the premises for the inclusion proof
– Marc
Oct 14, 2018 at 15:25
• @Marc that's the same because a point in $f[N_{\delta,X}(x)]$ is exactly of the form $f(x')$ for some $x' \in N_{\delta,X}(x)$. Oct 14, 2018 at 15:28
• But can you then use that for the inclusion proof? My head still can't really itself wrap around it, logically speaking. You have $$A: x'\in N_{\delta ,X}(x) \implies B: f(x')\in N_{\epsilon,Y}(f(x))$$ and you have $$A:x'\in N_{\delta,X}(x) \implies C:f(x')\in f[N_{\delta,X}(x)]$$ but to me these seem two separate logical statements: $A\implies B$ and $A\implies C$ but not $B\implies C$
– Marc
Oct 14, 2018 at 15:36
• @Marc There is nothing fancy going on at all. It's just a restatement of definitions. If we pick any poin $y$ in $f[N_{\delta,X}(x)]$ then $y = f(x')$ for some $x' \in N_{\delta,X}(x)$ and this means that $d_X(x,x') < \delta$ and we can apply uniform continuity for those two points and get $d_Y(f(x), f(x')) < \varepsilon$ which exactly means that $y = f(x') \in N_{\varepsilon, Y}(f(x))$. In uniform continuity we have two $\forall x$ quantifiers, in the neighbourhood statement we have too, but one is hidden in the inclusion, which is a statement of the form $\forall x \in A: x \in B$ as well. Oct 14, 2018 at 15:42
• @Marc The statement $f[A] \subseteq B$ can be logically written as $$\forall x \in A: f(x) \in B$$ and if you fill that in (1) and expand what it means to be in the ball we exactly get uniform continuity again. Oct 14, 2018 at 15:47