1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – William Elliot Apr 18 '17 at 1:38
  • 1
    $\begingroup$ Indeed, the closure of a path connected set need not be path connected. { (x, sin 1/x) | 0 < x } is the classic example. $\endgroup$ – William Elliot Apr 18 '17 at 1:46
2
$\begingroup$

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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.