# Collecting definitions of continuity.

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?

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

-
• 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.)
-
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