Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $X$ and $Y$ denote topological spaces and consider a function $f : X \rightarrow Y$. I'm collecting possible definitions/characterizations of the statement "$f$ is continuous."

Here's two to get us started.

  • (Preimages of Open Sets) $f$ is continuous iff for all open $B \subseteq Y$ it holds that $f^{-1}(B)$ is open in $X$.
  • (Direct Images of Open Neighbourhoods) $f$ is continuous iff for all $x \in X$ and all open $B \ni f(x)$, there exists open $A \ni x$ such that $f(A) \subseteq B$.

Bring 'em on!

Edit: Intuitively sensible definitions of continuity, which nevertheless fail, are also welcome (provide a counterexample).

Edit2: Does anyone know a way of characterising continuity in terms of boundary operators?

share|cite|improve this question
Preimages of closed sets. – P.. Mar 6 '13 at 20:45
Should this be community wiki & big list? – Robert Haraway Mar 6 '13 at 20:53
@RobertHaraway How do I community wikify it? – goblin Mar 6 '13 at 20:54
Maybe check Bourbaki. – Damien L Mar 6 '13 at 20:58

For every net $(x_i)_{i \in I}$ that converges to some $x$ in $X$, $f(x_i)_{i \in I}$ converges to $f(x)$.

For every filter $\mathcal{F}$ on $X$ that converges to $x \in X$, $f[\mathcal{F}]$ converges to $f(x)$.

The inverse image of a closed set in $Y$ is closed in $X$.

For a fixed base $\mathcal{B}$ of $Y$, the inverse image of $O$ is open for every $O \in \mathcal{B}$.

As the previous statement, but then for subbases.

For every $A \subset X$, $f[\overline{A}] \subset \overline{f[A]}$

For every $B \subset Y$, $\overline{f^{-1}[B]} \subset f^{-1}[\overline{B}]$

For every $B \subset Y$, $f^{-1}[\operatorname{Int}(B)] \subset \operatorname{Int}(f^{-1}[B])$

share|cite|improve this answer
  • Inverse images of closed sets are closed.
  • $f [ \overline{A} ] \subseteq \overline{ f[A] }$ for all $A \subseteq X$.
  • $\overline{f^{-1} [ B ] } \subseteq f^{-1} [ \overline{B} ]$ for all $B \subseteq Y$;
  • $f^{-1} [ \mathrm{Int} (B) ] \subseteq \mathrm{Int} ( f^{-1} [ B ] )$ for all $B \subseteq Y$.
  • whenever $\langle x_\sigma : \sigma \in \Sigma \rangle$ is a net in $X$ with limit $x$, then $f(x)$ is a limit of the net $\langle f(x_\sigma) : \sigma \in \Sigma \rangle$ in $Y$. (Note that for this I am not assuming that nets have unique limits.)
share|cite|improve this answer
Does the fifth dot point require qualification? As in "If $X$ and/or $Y$ is Hausdorff, then whenever..." – goblin Mar 6 '13 at 20:47
Also, does the fourth dot point have a "direct images" version? – goblin Mar 6 '13 at 20:48
Hausdorff is not needed for the nets characterization. If a net has 2 limits, then their images (if different) are also limits of the image net. – Henno Brandsma Mar 6 '13 at 20:54
@user18921: I'll clarify the fifth point, but in general I don't assume that nets have unique limits. As for the fourth point, the natural candidate for a "direct image" version would be $f [ \mathrm{Int} ( A ) ] \subseteq \mathrm{Int} ( f [ A ] )$, which fails for constant functions $\mathbb R \to \mathbb R$. – arjafi Mar 6 '13 at 20:58
In my opinion the natural candidate for a "direct image" version would be $\mathrm{int}(f(A))\subseteq f(\mathrm{int}(A))$ which is to the second dot point as the third is to the fourth. However, this fails, too. Take $X=\mathbb R$, $Y=\mathbb R_{\ge0}$, $f(x)=x^2$, and $A=\mathbb R_{\ge0}$. Then $\mathrm{int}(f(A))=Y$ but $f(\mathrm{int}(A))=\mathbb R_{>0}$. For bijective functions, however, it is just the dual of (2). – Stefan Hamcke Mar 6 '13 at 22:11

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.