Is the closure of path-connected set path connected?

If $U$ is a connected open set in $\mathbb R^n$, is $\bar U$ path-connected? (that is, for any two points $x_1,x_2$ in the closure of $U$, can we find a continuous path connecting them?)

Solution 1:

For a counterexample, take $$U=\{(x,y)\in\mathbb{R}^2\colon \lvert y-\sin(x^{-1})\rvert<x\}.$$ Its closure will be $$\overline U=\{(x,y)\in\mathbb{R}^2\colon \lvert y-\sin(x^{-1})\rvert\le x\} \cup\bigl(\{0\}\times[-1,1]\bigr). $$ No curve from the interior can reach the vertical segment along the $y$ axis.

Solution 2:

I fill in the details of Olivier Bégassat's proposal: Let $Y$ be the $y$-axis, and let $X$ be the union of the open vertical segments $L_x$ centered at the points $p_x=\bigl(x,\sin(1/x)\bigr)$ with $x>0$, with length $2x$. Now I invoke the "comb theorem" (patent pending ;-) ): If $W$ is a topological space, and $Z$ (the "shaft")and $Z_i, i\in I$ (the "teeth") are connected (resp., path-connected) subsets of $W$ such that $Z_i\cap Z\ne\emptyset$ for each $i$, then the union $Z\cup\bigl(\cup_i Z_i\bigr)$ is a connected (resp., path-connected) set.

In our case of interest the shaft $Z$ is the graph of the function $y=\sin(1/x)$ for $x>0$, which is path-connected (the image of the path-connected set $(0,\infty)$ via the continuous function $\sin(1/x)\ $), and the teeth are the segments $L_x$, which obviously are path-connected. Therefore $X$ is path-connected.

The mapping $h:(0,\infty)\times(-1,1)\to X$ given by $h(x,t)=\bigl(x,\sin(1/x)+tx\bigr)$ is a homeomorphism, so $X$ is an open set. It remains to show that $\overline X$ is not path-connected. For this, suppose that some continuous path $\gamma:[0,1]\to\overline X$ is such that $\gamma(0)=(0,0)$ and $\gamma(1)=(1/\pi,0)$. Let $$c=\sup\{x\in[0,1]: \gamma(x)\in Y\}\,.$$ Since $Y$ is closed, then we actually have $\gamma(c)\in Y$, and so $\gamma\bigl([c,1]\bigr)\cap Y=\emptyset$.

Let $(x,y)\in\overline X$, with $x>0$. Then some sequence $(x_n,y_n)\in X$ converges to $(x,y)$, and so $y_n=\sin(1/x_n)+t_nx_n$ with $t_n\in(-1,1)$. Taking a subsequence, we can suppose that $t_n\to t\in[-1,1]$, which implies $y=\sin(1/x)+tx$. This shows that the points of $\overline X$ outside $Y$ are in the union, say $T$, of the closed segments $\overline{L_x}$ with $x>0$ (which is visually obvious). In particular we have $\gamma\bigl((c,1]\bigr)\subseteq T$.

Let $\gamma(x)=\bigl(\alpha(x),\beta(x)\bigr)$. If $c<d\leq1$, then $\alpha(d)>0$. Let $k$ be a positive integer such that $s,t<\min\{1/2,\alpha(d)\}$, where

$$s=\frac1{2k\pi+\frac\pi2},\ t=\frac1{2k\pi-\frac\pi2}.$$

Since $\alpha$ is continuous, then some $d^\prime,d^{\prime\prime}\in(c,d)$ satisfy $\alpha(d^\prime)=s$ and $\alpha(d^{\prime\prime})=t$. In this case we have $|\beta(d^\prime)-\sin(1/s)|=|\beta(d^\prime)-1|\leq s\leq1/2$, and so $\beta(d^\prime)\geq1/2$. Similarly we have $|\beta(d^{\prime\prime})-\sin(1/t)|=|\beta(d^{\prime\prime})+1|\leq t\leq1/2$, which implies $\beta(d^{\prime\prime})\leq-1/2$. Since $\beta$ is continuous, then $\beta(d^{\prime\prime\prime})=1/2$ for some $d^{\prime\prime\prime}$ between $d^\prime$ and $d^{\prime\prime}$.

Consequently, as $d$ goes to $c$, the points $\gamma(d^{\prime\prime\prime})$ converge to $(0,1/2)$. Therefore $(0,1/2)\in\overline{\gamma\bigl([c,1]\bigr)}$, which is absurd because $\gamma\bigl([c,1]\bigr)$ is compact, hence closed, and we already see that $\gamma\bigl([c,1]\bigr)\cap Y=\emptyset$. This contradiction shows that $\overline X$ is not path-connected.