Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When we define continuity using open balls, we define $$\forall \epsilon >0, \exists\delta>0:f(B_\delta(a))\subset B_\epsilon(f(a))$$ Let us consider everything unspecified to be in $\mathbb{R^n}$ i.e. $f:\mathbb{R^n}\rightarrow \mathbb{R^n}$and usual distance functions if these are necessary to be specified.

What is so special about open balls here? It seems closed balls work fine too.

share|cite|improve this question
Did you studied topology? Im asking because you are using this tag. – Tomás Jan 12 '13 at 11:07
In $\mathbb R^n$ you're so lucky that closed balls are just closures of open balls. But there are plenty of metric spaces where that's not the case. For example $\mathbb N$ with the standard discrete metrix $d(x,y) = \delta_{xy} = \begin{cases}1 & \text{if } x\neq y \cr 0 & \text{if } x=y \end{cases}$. In that space we have that the open $1$-ball around $n\in \mathbb N$ is just $\{n\}$, but the closed $1$-ball is all of $\mathbb N$. – kahen Jan 12 '13 at 11:17
up vote 3 down vote accepted

Let $(X,d),(Y,\rho)$ be metric spaces and $f:X\to Y$. For the rest of the post we will say $f$ is Cont1 at $a\in X$ if $$\forall \epsilon>0\exists \delta>0:f(B_{\delta}(a))\subseteq B_{\epsilon}(f(a))$$ and Cont2 if $$\forall \epsilon>0\exists \delta>0:f(\overline{B}_{\delta}(a))\subseteq \overline{B}_{\epsilon}(f(a))$$

We will prove that $f$ is Cont1 $\iff$ $f$ is Cont2.

Cont1$\implies$ Cont2:

Let $\epsilon>0$. Then,

$$\exists \delta>0:f(B_{\delta}(a))\subseteq B_{\epsilon}(f(a))$$ Let $0<\delta^{\prime}<\delta$. Then, $$f(\overline{B}_{\delta^{\prime}}(a))\subseteq f(B_{\delta}(a))\subseteq B_{\epsilon}(f(a))\subseteq \overline{B}_{\epsilon}(f(a))$$ which is Cont2

Cont2$\implies$ Cont1:

Let $\epsilon>0$ and $0<\epsilon^{\prime}<\epsilon$. Then,

$$\exists \delta>0:f(\overline{B}_{\delta}(a))\subseteq \overline{B}_{\epsilon^{\prime}}(f(a))$$ Then, $$f(B_{\delta}(a))\subseteq f(\overline{B}_{\delta}(a))\subseteq \overline{B}_{\epsilon^{\prime}}(f(a))\subseteq B_{\epsilon}(f(a))$$ which is Cont1

Therefore whether you define the balls in the definition of continuity to be open or closed makes no difference

share|cite|improve this answer
Small MathJax/(La)TeX tip: There's no need to write ^\prime. Just using a regular old apostrophe will work just as well: Compare $a^{\prime}$ to $a'$ (right click on the math to view the MathJax source). – kahen Jan 12 '13 at 11:26
@kahen I didn't know that. Thank you very very much for this tip'. – Nameless Jan 12 '13 at 11:28

As long as your require the radius of the ball to be strictly positive note that closed balls contain open balls, so the requirement for "open balls" is satisfied.

Generally, however, the topological definition for a continuous map is "the preimage of an open set is open". It is enough to require this for basic open sets, which in metric spaces corresponds to open balls. Equally, however, one can require this for closed sets as well, namely the preimage of a closed set is closed. Do note that in this case the notion is not transferred immediately to closed balls, there are closed sets which are not the union/intersection of closed balls (for example the $x$-axis in the plane).

share|cite|improve this answer
The metric $\epsilon$-$\delta$ definition of continuity is "local", i.e. it only speaks of continuity at a point in contrast to the topological definition. But there is of course also a topological notion of continuity of a function $f:X\to Y$ at a point $x \in X$: For all neighbourhoods $V$ of $f(x)$ there exists a neighbourhood $U$ of $x$ such that $f(U) \subseteq V$. And $f$ is continuous iff it's continuous at each $x\in X$. – kahen Jan 12 '13 at 11:14
@WantTobeAbstract: "it only speaks of continuity at a point", what's more to add? – Asaf Karagila Sep 19 '14 at 12:13
@WantTobeAbstract: (1) You can ask a new question, instead of repeatedly pinging someone in the comments. This amounts to harassment, and is frowned upon here. (2) The term local was qualified, and the idea is that it happens at a single point and not everywhere. – Asaf Karagila Sep 19 '14 at 12:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.