# Is the space path connected?

Let $$X=\bigcup_{n\in \mathbb N} \bigg( \{1/n\} \times \mathbb{R}\cup\mathbb{R} \times \{1/n\}\bigg)\cup \{(0,0)\}.$$ I want to check if this is a path connected space. I can see that $0$ is in the closure of space $X\setminus\{(0,0)\}$, Which seems to be path connected. But I know that the closure of a path connected space is not necessarily path connected.

• No. {(0,0)} can be separated from the rest of the space no matter what the value of n is. Likely your description of the intended is incomplete. – William Elliot Apr 18 '17 at 1:38
• Indeed, the closure of a path connected set need not be path connected. { (x, sin 1/x) | 0 < x } is the classic example. – William Elliot Apr 18 '17 at 1:46

Let $j,k\in\mathbb N$ and fix $x\in\{1/j\}\times\mathbb R$ and $y\in\mathbb R\times\{1/k\}$. Define $p=(1/j,1/k)$. Surely, $p\in X$. Since $x$ and $p$ are contained in the line $\{1/j\}\times\mathbb R$, there is a path joining $x$ to $p$. Similarly, we conclude that there is a path joining $y$ to $p$. Therefore, we are able to construct a path joining $x$ to $y$. This way we proved that $X\setminus\{(0,0)\}$ is path-connected.

To prove that $X$ is path-connected, it only lasts to prove that $(0,0)$ is path-connected to any $x\in X\setminus\{(0,0)\}$.

Let $x\in\{1/k\}\times \mathbb R$ and we will construct below a path $\gamma$ joining $x$ to $(0,0)$.

Pick a path $\gamma_0:[0,1]\to X$ such that $\gamma_0(0)=x$ and $\gamma_0(1)=(1/k,1/k)$.

We can choose a path $\gamma_1:[0,1]\to X$ such that $\gamma_1(0)=(1/k,1/k)$, $\gamma_1(1)=(1/(k+1),1/(k+1))$ and $$Im \gamma_1\subset[1/(k+1),1/k]\times [1/(k+1),1/k],$$ (it's just a matter of concatenating some sides of the square)

For $j\geq2$, we do it analogously and pick a path $\gamma_j:[0,1]\to X$ such that $\gamma_j(0)=(1/(k+j-1),1/(k+j-1))$, $\gamma_j(1)=(1/(k+j),1/(k+j))$ and $$Im \gamma_j\subset[1/(k+j),1/(k+j-1)]\times [1/(k+j),1/(k+j-1)].$$

This way we get a sequence of paths $\gamma_0$, $\gamma_1$,..., $\gamma_j$,... such that $\gamma_{j-1}(1)=\gamma_j(0)$, for each $j\geq1$. Then, for each $n$ we can concatenate $\gamma_0$, $\gamma_1$,...,$\gamma_n$ and get a path joining $x$ to $(1/(k+n),1/(k+n))$. This way we could get as close to $0$ as we wish but still can't reach $0$.

To reach $0$, we must do a kind of infinite concatenation. Define $\gamma:[0,1]\to X$ such that $$\gamma(t)= \left\{\begin{array}{l}\gamma_j(2^j(t-(1-2^{-j}))), \mbox{ if } t\in[1-2^{-j},1-2^{-(j+1)})\\ 0 ,\mbox{ if } t=1\end{array}\right..$$ What we are doing above is putting each $\gamma_j$ to be the image of $\gamma$ in the interval $[1-2^{-j},1-2^{-(j+1)}]$.

Now it only lasts to prove that $\gamma$ is continuous. The continuity at the points $t\in[0,1)$ follows from the continuity of the $\gamma_j$'s and the fact that $\gamma_j(1)=\gamma_{j+1}(0)$ for all $j\geq0$. The continuity at $1$ follows from the fact below.

Fix $j\geq0$. If $t\in[1-2^{-j},1)$, then $t\in(1-2^{-j_0},1-2^{-(j_0+1)}]$ for some $j_0\geq j$. Therefore,

\begin{align}\gamma(t)\in Im \gamma_{j_0}&\subset[1/(k+j_0),1/(k+j_0-1)]\times [1/(k+j_0),1/(k+j_0-1)]\\&\subset [0,1/(k+j_0-1)]\times[0,1/(k+j_0-1)]\\&\subset [0,1/(k+j-1)]\times[0,1/(k+j-1)]. \end{align}