Is this “Combesque” Space path connected.

Question

Pardon my English.

I have to prove that this space is or isn’t path connected:$$K=\bigl\{(0,0)\bigr\}\cup\bigl[(0,2)\times(0,2)\bigr]\setminus\left\{\left(\frac1n,y\right)\,\middle|\,y\in(0,1)\right\}.$$

We have figured out that it is connected, and kind of similar to a comb space.

How would one go about doing that?

By finding, for any two points in $K$, an explicit path connecting them. — Robert Shore, Aug 16 '21 at 17:30
@freakish I'm pretty confident $n$ is intended to range over the positive integers. But you're right that should be explicit in the problem statement. — Robert Shore, Aug 16 '21 at 17:32
@freakish yes it was indeed n\in\mathbb{N} and thank you so much for your help. Your solution is even more detailed than expected! — Xelamoon, Aug 16 '21 at 19:57

freakish · Accepted Answer · 2021-08-16T18:14:50.303

I assume the last piece of your $K$ is actually $\{(1/n, y)\ |\ n\in\mathbb{N}, y\in(0,1)\}$. In that case your $K$ is not path connected. So you have an example of a connected but not path connected space.

Lemma 1. If $\lambda:[0,1]\to X$ is a path to a Hausdorff space then the image is locally connected.

Proof. Since $X$ is Hausdorff and $[0,1]$ is compact, then $\lambda$ is closed, in particular a quotient map onto its image. And being locally connected is preserved under quotients. $\Box$

Lemma 2. If $F\subseteq K$ is any subset such that $(0,0)\in F$ and $(0,0)$ is not isolated in $F$ then $F$ is not locally connected.

Proof. Consider the open ball $B$ around $(0,0)$ of radius $1$ (the radius is important so that we don't go above vertical lines). Now let $V\subseteq B\cap F$ be any open neighbourhood of $(0,0)$. Since $(0,0)$ is not isolated in $F$ then $V$ contains some point $v\neq (0,0)$. Write down $v=(x_v,y_v)$ and note that $x_v>0$ by our definition of $K$. Let $n\in\mathbb{N}$ be such that $0<1/n<x_v$. Since $V$ is a subset of $B$ and $K$ then no point in $V$ can have the first coordinate equal to $1/n$. This implies that $V$ can be decomposed into two nonempty, open and disjoint subsets:

$$V^-=\{(x,y)\in V\ |\ x<1/n\}$$ $$V^+=\{(x,y)\in V\ |\ x>1/n\}$$ $$V=V^-\cup V^+$$

In particular $V$ is disconnected. By the arbitrary choice of $V$ we conclude that $F$ is not locally connected. $\Box$

Corollary. $K$ is not path connected.

Proof. Assume $\lambda:[0,1]\to K$ is a path, $F=im(\lambda)$ and $(0,0)\in F$. Then by Lemma 1 our $F$ is locally connected. But then by Lemma 2 our $(0,0)$ point has to be isolated in $F$. In particular either $F$ is disconnected or $F=\{(0,0)\}$. The first case of course cannot happen because $F$ is an image of a connected space. Thus $F=\{(0,0)\}$ and so $\lambda$ is a constant path. In particular no point other than $(0,0)$ can be connected to $(0,0)$ inside $K$. $\Box$

Is this “Combesque” Space path connected.

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