\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
deepdream of
          a sidewalk
Show Answer
\( \newcommand{\cat}[1] {\mathrm{#1}} \newcommand{\catobj}[1] {\operatorname{Obj}(\mathrm{#1})} \newcommand{\cathom}[1] {\operatorname{Hom}_{\cat{#1}}} \newcommand{\multiBetaReduction}[0] {\twoheadrightarrow_{\beta}} \newcommand{\betaReduction}[0] {\rightarrow_{\beta}} \newcommand{\betaEq}[0] {=_{\beta}} \newcommand{\string}[1] {\texttt{"}\mathtt{#1}\texttt{"}} \newcommand{\symbolq}[1] {\texttt{`}\mathtt{#1}\texttt{'}} \newcommand{\groupMul}[1] { \cdot_{\small{#1}}} \newcommand{\inv}[1] {#1^{-1} } \newcommand{\bm}[1] { \boldsymbol{#1} } \require{physics} \require{ams} \)
Math and science::Topology

Continuous maps

Continuous maps

Let \( X \) and \( Y \) be topological spaces. A function \( f : X \to Y \) is continuous iff [for every something, that something has some property].

In short, continuity means that [phrased in a few simple words...].

Some results

Continuous maps preserve convergence of sequences.

Let \( f : X \to Y \) be a continuous map, and let \( (x_n) \) be a sequence in \( X \) converging to \( x \in X \); then the sequence [...] converges to [...].

In metric spaces this lemma is an if and only if statement, whereas for topological spaces we are restricted to only the forward implication above; it is possible to construct discontinuous maps of topological spaces that, nevertheless, preserve convergence of sequences.

The composite of continuous maps [is always continuous/need not be continuous?].

The inverse of a continuous bijection [is always continuous/need not be continuous?].

Munkres presents three statements that are equivalent to stating that a function is continuous:

Continuity equivalences

Let \( X \) and \( Y \) be topological spaces and let \( f : X \to Y \) be a function. The the following are equivalent:

  1. \( f \) is continuous.
  2. For every subset \( A \) of \( X \), one has [\( f(\bar{A}) \subseteq \text{what set?} \)].
  3. For every closed set \( B \) of \( Y \), the set [...] is [...].
  4. For each \( x \in X \) and each neighbourhood \( V \) of \( f(x) \), there is a [...] such that [...].