Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The question is pretty much in the title, I'm looking for an example of a locally connected space and continuous mapping such that the image is not locally connected.


EDIT: Corrected the phrasing to the intended meaning.

share|improve this question
Did you want some condition on the image of the map? Otherwise, pick any non-locally connected space and any map from a single point to this space. –  Daniel Rust Jun 25 '13 at 15:40
The identity from $\Bbb R$ with the discrete topology to the Sorgenfry Line. –  David Mitra Jun 25 '13 at 15:42
@James: The image $f(U)$ need not be open unless the map $f$ is an open mapping. –  Stefan Hamcke Jun 25 '13 at 15:46
See my answer to the question Is the image of a path or arc locally path/arc connected?. It shows that even the continuous image of the unit interval need not be locally connected. –  Stefan Hamcke Jun 25 '13 at 15:53
May be the following will work. Let $X=[0;1]$ and $f:X\to\mathbb C$ such that $f(x)=xe^{i/x}$ for each $x\in X$. –  Alex Ravsky Jun 25 '13 at 15:55

3 Answers 3

up vote 1 down vote accepted

Consider the following variant on the topologist's sine curve.

enter image description here

This space $X$ consists of the graph of $y = \sin(\pi/x)$ for $0<x<1$, together with a closed arc from the point $(1,0)$ to $(0,0)$. Note that $X$ is not locally connected at $(0,0)$.

However, there exists a continuous surjection $f\colon [0,2)\to X$. Specifically, $f(0) = (0,0)$ and $f(1) = (1,0)$, with $f(t)$ following along the bottom curve for $0\leq t\leq 1$. For $t>1$, the function follows along the sine curve, i.e. $$ f(t) \;=\; \left(2-t,\sin\left(\frac{\pi}{2-t}\right)\right)\qquad\text{for }t> 1. $$

share|improve this answer
Another nice example. I really liked that graph, is there a simple way to create such graphs? :) –  Serpahimz Jun 25 '13 at 17:23
I borrowed the picture from a previous answer of mine. I don't recall how I made it, but I would guess Mathematica. –  Jim Belk Jun 25 '13 at 18:03

Boring example: any space $X$ is the continuous image of the discrete topology on $X$ (using the identity and noting that any function with a discrete domain is continuous). A discrete space is trivially locally connected (all singleton sets). Now let $X$ be any non-locally connected space.

share|improve this answer

The graph-parametrisation of the graph of $\sin \frac{1}{x}$ for $x>0$ ? Instead of letting the graph trail off to the right as $x\to\infty$, just turn it around and let the curve run along the interval $[-1,1]$ on the $y$-axis. Then every point of this interval will have the property that local connectivity fails.

share|improve this answer
Maybe this is more reasonable as a comment? –  Daniel Rust Jun 25 '13 at 15:54
What? That graph is locally connected. –  Chris Eagle Jun 25 '13 at 15:56
It seems that the graph of $\sin \frac{1}{x}$ for $x>0$ is homeomorhic to $\mathbb R$. –  Alex Ravsky Jun 25 '13 at 15:57
I added the part of the curve along the $y$-axis. –  user72694 Jun 26 '13 at 7:17

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.