# Local Connectedness is not Preserved under Continuous Mapping

Jump to navigation Jump to search

## Theorem

Let $\struct {A, \tau_1}$ and $\struct {B, \tau_2}$ be topological spaces.

Let $f: A \to B$ be a continuous mapping.

Let $\struct {\Img f, \tau_3}$ be the image of $f$ with the subspace topology of $B$.

Let $\struct {A, \tau_1}$ be locally connected.

Then it is not necessarily the case that $\struct {\Img f, \tau_3}$ is also locally connected.

## Proof

Let $\struct {A, \tau_1}$ be any countable discrete space.

Let $B \subseteq \R$ be the set of all points on $\R$ defined as:

$B := \set 0 \cup \set {\dfrac 1 n: n \in \Z_{>0} }$

Let $\struct {B, \tau_2}$ be the integer reciprocal space with zero under the usual (Euclidean) topology.

From Discrete Space is Locally Connected, $\struct {A, \tau_1}$ is locally connected.

Let $f: A \to B$ be a bijection.

From Mapping from Discrete Space is Continuous, $f: A \to B$ is a continuous mapping.

From Integer Reciprocal Space with Zero is not Locally Connected, $\struct {B, \tau_2} = \struct {\Img f, \tau_3}$ is not locally connected.

$\blacksquare$