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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?

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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.

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