# Why is the deleted comb space still connected?

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...

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

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.