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?


1 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.

  • $\begingroup$ 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 $\endgroup$
    – Marc
    Oct 14, 2018 at 15:25
  • $\begingroup$ @Marc that's the same because a point in $f[N_{\delta,X}(x)]$ is exactly of the form $f(x')$ for some $x' \in N_{\delta,X}(x)$. $\endgroup$ Oct 14, 2018 at 15:28
  • $\begingroup$ 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 $B\implies C$ $\endgroup$
    – Marc
    Oct 14, 2018 at 15:36
  • $\begingroup$ @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. $\endgroup$ Oct 14, 2018 at 15:42
  • $\begingroup$ @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. $\endgroup$ Oct 14, 2018 at 15:47

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