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


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 $x\in X$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.