# Comb space: cannot understand why it is not locally path-connected

Wikipedia says that the comb space is path-connected but not locally path-connected. I cannot see this. Even when I read here that at any point in $\{0\}\times (0,1]$ it is not locally path-connected, I still cannot see it. For example, if we took the point $p=(0,1)$, surely the neighbourhood $\{0\}\times (1-\epsilon,1]$ is open and path-connected, and contains $p$, and if we make $\epsilon$ small enough we can get it to fit inside any open neighbourhood of $p$.

• That isn't an open set. Since the comb space has the subspace topology inherited from $\mathbb{R}^2$, you should intersect open balls with your comb space to see what open sets look like in it. Once you draw a small enough open ball around a point in $\{0\}\times (0,1]$, you should see that this isn't path connected (or even connected). – Hayden May 2 '17 at 16:21

The set $A:=\{0\}\times (1-\epsilon,1]$ is not open in the comb space $C$. Indeed $((1/n, 1))_{n\in\mathbb N}$ is a sequence in $C\setminus A$ which converges to $(0,1)\in A$.
• I'm a bit confused: if I take a point $x=(0,x_0)\in \{0\}\times (0, 1]$ and create a ball of radius $\epsilon<x_0$ centered at $x$, denoted $B_{\epsilon}(x)$, it will intersect infinitely many of the comb teeth to its right in "open intervals." Local path connectedness requires that there's a path-connected open subset of this neighborhood. So couldn't I pick some $T=\{\frac{1}{n}\} \times (0,1]$ that intersects $B_{\epsilon}(x)$ and take a look at its interval? I can construct the interval by intersecting open balls of radius $\epsilon'<\frac{1}{n}-\frac{1}{n+1}$ with $T$, so it's open. – user413923 Sep 29 '19 at 21:15
• @Daniel check the definition again, we are looking for a small path-connected neighbourhood of $x$, I.e. the set must contain $x$. – Carsten S Sep 29 '19 at 23:38