1
$\begingroup$

I read from wikipedia that the deleted comb space:

$$ (\{0\} \times \{0,1\}) \bigcup (K \times [0,1]) \bigcup ([0,1] \times \{ 0 \}) $$

has a well known result of being connected.. But how is it connected if there's a separate point $p=<0,1>$? Or is it because since $p$ is a point and is not considered to be among the open sets of $\mathbb{R}$ which are in the form of intervals...

$\endgroup$
  • 1
    $\begingroup$ The title sounds like the answer is a caching problem. $\endgroup$ – Asaf Karagila Jun 13 '19 at 8:59
1
$\begingroup$

Clearly, if $Y=\bigl(K\times[0,1]\bigr)\cup\bigl([0,1]\times\{0\}\bigr)$, then $Y$ is path connected, and therefore connected. But your set lies between $Y$ is its closure. Therefore, it is connected too.

$\endgroup$

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.