# Deleted comb space modified to be locally connected at zero

Let $$D$$ refer to the deleted comb space. I read that if we add to $$D$$ all points of the form $$\{0\} \times \{1/n\}$$ for $$n \in \mathbb{Z}_+$$, the origin becomes locally connected but not locally path connected.

I dont understand this... I read that the original comb space is not locally connected despite already having the points $$\{0\} \times [0,1]$$. So how is this different?

The comb space $$C$$ is not locally connected at all of its points. In fact, one can show that it is locally connected precisely at the points of $$C \setminus \{0\}\times (0,1]$$.
The claim here is that $$D' = D \cup \{(0,1/n) \mid n \in \mathbb Z \}$$ is locally connected at $$(0,0)$$ but not locally path connected at $$(0,0)$$. The second part is obvious: No neighborhood of $$(0,0)$$ in $$D'$$ is path connected because there does not exist a path from $$(0,0)$$ to any $$(0,1/n)$$. See also Deleted comb space is not path connected.
To see that $$D'$$ is locally connected at $$(0,0)$$, let $$U$$ be any neighborhood of $$(0,0)$$ in $$D'$$. Choose $$r > 0$$ such that $$V = \left( [0,r) \times [0,r) \right) \cap D' \subset U$$. Then $$V$$ is a connected open neighborhood of $$(0,0)$$. This comes from the fact that $$W = \left( (0,r) \times [0,r) \right) \cap D' = \left( (0,r) \times [0,r) \right) \cap C$$ is connected. Hence the closure $$\overline{W}^C$$ in $$C$$ is connected and since we have $$W \subset V \subset \overline{W}^C$$, also $$V$$ is connected.