This is a problem I cannot prove,

Show that if $X$ is connected and is such that for every $x \in X$ there exists and open path-connected set $U$ with $x \in U$ then it is path-connected.

Here's my poor attempt

$X$ is connecetd so let some continuous function $f:X \to D$ to a discrete space exists. Then, this map is constant. Since $U$ is path-connecetd, it is also connected. Therefore, any restrcition $f_U=f$. i.e. The map is constant too.

Let $x_1,x_2 \in U$. Since it is path-connected, there exists some path $\alpha:I \to U$. Then, $f(x_1)=f(x_2)=f(\alpha(0))=f(\alpha(1)) \in D$. Since $f$ is continuous, and noting that a point in the discrete space is open by the discrete topology, $f^{-1}\{f(\alpha(t)\} \in X$ is open. In fact, this is the entire set $X$.

Now, I'm thinking that having elements/points from path-connecetd $U$ now related to the entire set $X$ has something to do with...showing $X$ is also path-connected. Or something. But I can't rigorously proceed from here.

Or maybe my entire thought process is wrong. I don't know what to do with the vast amount of information, connectedness, open subsets, path-connected subset, continuous map to a discrete space is constant etc etc.

Which ones I can use, which ones I don't need is just to complicated for me to guess. Please help


1 Answer 1


HINT: For $x\in X$ let $C(x)$ be the path component of $X$.

  • Show that $C(x)$ is open for each $x\in X$. Here’s where you’ll use the hypothesis that each point has a path-connected open nbhd.
  • Show that $\{C(x):x\in X\}$ is a partition of $X$.
  • Use the connectedness of $X$ to conclude that $X=C(x)$ for each $x\in X$.

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