Continuity of a map of topological spaces at a point and globally

Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $f : X \to Y$ be a function and let $x \in X$. Neighbourhoods are as in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open.

$f$ is continuous at $x$ if for every neighbourhood $V$ of $f(x)$ in $Y$ the preimage $f^{-1}[V]$ is a neighbourhood of $x$ in $X$.

$f$ is continuous if it is continuous at every point of $X$.

The same condition with open sets only. $f$ is continuous at $x$ if and only if for every open $V \subseteq Y$ with $f(x) \in V$ there is an open $U \subseteq X$ with $x \in U$ and $f[U] \subseteq V$. Indeed, if $f$ is continuous at $x$ and $V$ is such an open set, then $V$ is a neighbourhood of $f(x)$, so $f^{-1}[V]$ is a neighbourhood of $x$ and contains an open $U \ni x$, which satisfies $f[U] \subseteq V$. Conversely, given the displayed condition and a neighbourhood $V$ of $f(x)$, fix open $V_0$ with $f(x) \in V_0 \subseteq V$ and then open $U \ni x$ with $f[U] \subseteq V_0$; then $x \in U \subseteq f^{-1}[V_0] \subseteq f^{-1}[V]$, so $f^{-1}[V]$ is a neighbourhood of $x$. Both forms are used below and are the same statement written twice.

Preimage, not image. $f^{-1}[V] = \{\, x \in X : f(x) \in V \,\}$ is the preimage in the sense of Injection, surjection, bijection and is defined for every function, invertible or not; no inverse function is being asserted to exist. Continuity is a condition on preimages throughout, and the corresponding conditions on images define the open and closed maps of a later item, which are different notions.

• This is the metric definition when both topologies are metric topologies. For metric spaces, $\varepsilon$-$\delta$ continuity at $a$ (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form) says that every ball around $f(a)$ has a ball around $a$ mapped into it, and the balls around a point are a neighbourhood base there; the identification is carried out where metrizable spaces are defined later on this page. Nothing about a metric survives in the definition above: continuity is a relation between two topologies and a function, and it is meaningless to ask whether a function between bare sets is continuous.

• Continuity depends on both topologies, and coarsening the target or refining the source only helps. If $f$ is continuous and $\mathcal{T}_X$ is replaced by a finer topology, or $\mathcal{T}_Y$ by a coarser one, $f$ remains continuous, since each condition to be verified is weakened and each neighbourhood available in the source is still available. In particular every map out of a discrete space and every map into an indiscrete space is continuous (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

• Continuity at a point is strictly weaker than continuity. A function may be continuous at exactly one point, and the definition above is deliberately local so that the sequential criteria proved later can be stated pointwise.