$\mathbf{Attempt:}$Consider the open cover $\bigcup_{x\in S} B(x,r_x)$ of $S$ with each $B(x,r_x)\subset S$ for some $r_x>0$.

We can invoke the Lindelöf property to get ourselves a countable subcover $\bigcup_{i\in \mathbb{N}} B(x_i,r_{x_i})$ of $\bigcup_{x\in S} B(x,r_x)$ which still covers $S$. We know that every ball $B(x,r_x)=\{y\in \mathbb{R}^n: d(x,y)<r_x\}\subset S$ is path connected.

$\underline{Claim:}$ The union over a family of non-empty path-connected sets $\{A_i\}_{i\in I_n}$ remains path-connected if for $k =1,2,...,n-1$

$\displaystyle(\bigcup_{i\in I_n\setminus J_k}A_i)\bigcap (\bigcup_{t\in J_k} A_t)\neq \emptyset$

$\forall n \geq 2$, where $I_n=\{1,2,...,n\}$ and $J_k=\{s_1,s_2,...,s_k\}\subset I_n$ where $s_i$ takes arbitrary value and $s_i \neq s_j$ for any two $i\neq j$.

$\underline{Justification:} $ For $n=2$, consider $A_1$ and $A_2$ to be two path connected sets with non-empty intersection. Pick any two $a,b \in A_1 \cup A_2$. For $a,b \in A_1$ or $a, b \in A_2$ we can find readily find a path connecting them within that set. Now, for $a \in A_1$ and $b\in A_2$ (WLOG), define $h:[0,1] \to A_1\cup A_2$ such that $h(t)=f(2t)$ for $t \in [0,1/2]$ and $h(t)=g(2t-1)$ for $t \in [1/2,1]$ where $f:[0,1] \to A_1$ and $g:[0,1] \to A_2$ such that $f(0)=a$ and $f(1)=c$ and $g(0)=c$ and $g(1)=b$, $c \in A_1\cap A_2$. Clearly, $h$ is continuous on $[0,1]$ with $h(0)=a$ and $h(1)=b$. Hence $A_1 \cup A_2$ is path connected.

Let for $n=m$, $\displaystyle(\bigcup_{i\in I_m\setminus J_k}A_i)\bigcap (\bigcup_{t\in J_k} A_t)\neq \emptyset$.

Put $J_1=1$. Then we get at least one $r_1\neq 1$ such that $A_1\cap A_{r_1}\neq \emptyset$, i.e $A_1 \cup A_{r_1}$ is path connected. Next, let us set $J_2=\{1,r_1\}$. We get $r_2\neq 1, r_1$ with $A_{r_2}\cap(A_1\cup A_{r_1}) \neq \emptyset$. So $A_1\cup A_{r_1}\cup A_{r_2}$ is path connected.

Iterating this process until we exhaust all of $I_m$, we find that $\bigcup_{I_m} A_i$ remains path-connected. This argument holds for $m$, $m+1$ $\cdots$. So for any such countable family, it holds good as well.

Going back to our countable open cover of $S$, we must have $\displaystyle(\bigcup_{i\in \mathbb{N}\setminus J_k}B(x_i,r_{x_i}))\bigcap (\bigcup_{t \in J_k} B(x_t,r_{x_t}))\neq \emptyset$, for all $k\in \mathbb{N}$ and any $J_k$. [Otherwise its emptiness would give us $C=\bigcup_{i\in \mathbb{N}\setminus J_k}B(x_i,r_{x_i})$ and $D=\bigcup_{t\in J_k}B(x_t,r_{x_t})$ for some $J_k$ corresponding to some $k$, such that $C\cap D=\emptyset$ and $C\cup D = S$, both of which are open in the subspace metric on $S$, implying that $S$ is disconnected].

We can now establish the path-connectedness of $S\subset \mathbb{R}^n$ by appealing to our pre-verified claim and choosing $A_i=B(x_i,r_{x_i})$.

$\text{Is this okay? Kindly Verify}.$

  • 1
    $\begingroup$ This proof looks fine to me. $\endgroup$ Dec 20 '20 at 23:54

I didn't read the whole proof, but it seems like it could work, but it is definitely not the "best" way to show the desired result, it just uses way too much machinery, using Lindelof's theorem and the way of showing that the thing you're constructing is path connected.

I'm not sure if you've seen the definition of locally path connected, but it reads:

$X$ is locally path connected if for all $p \in X$, for every open $U$ s.t. $x \in U$ there exists $V \subseteq U$ s.t. $V$ is open, $x \in V$, and $V$ is path connected.

Now you should be able to easily convince yourself that $\mathbb{R}^n$ is locally path connected. Then convince yourself that open subsets of locally path connected spaces are locally path connected in the subspace topology. So now if $U \subseteq \mathbb{R}^n$ is open, then $U$ is locally path connected. Then suppose (as in your problem) that $U$ is also connected, we want to show it is path connected. Here is an example of a proof that connected and locally path connected spaces are path connected, which will finish the proof of your result. Note that the result that connected, locally path connected spaces are path connected is (almost) immediate from the result that path components of locally path connected spaces are open, which isn't hard to show, so you could also show it that way.

  • $\begingroup$ This looks way better. Thank you for this one! $\endgroup$ Dec 21 '20 at 0:09
  • $\begingroup$ @SubhasisBiswas remember to accept the answer if the question's been resolved. $\endgroup$ Dec 21 '20 at 2:13

I will just streamline your proof, following your idea: suppose $S$ is open, nonempty and connected. Pick $p\in S$ and let $U$ be the set of points in $S$ that can be joined to $p$ by a path in $S$ and let $V$ be the complement of $U$. Then $S=U\cup V.$ Note $U$ is nonempty because it contains $p$.

Let $q\in U$ so that there is a path $\gamma$ joining $q$ to $p$. Now let $B$ be any ball in $S$ containing $q.\ B$ is convex so the path $\gamma': t\mapsto (1-t)q'+qt$ joins any point $q'\in B$ to $q.$ So $\gamma'*\gamma$ is then a path that joins $q'$ to $p$. Therefore $q\in B\subseteq U$ and so $U$ is open.

Now take $q\in V.$ Since $q$ cannot be joined to $p$ by a path and since the points from every ball $B$ about $q$ can be joined to $q$ by a path, it follows that any ball in $S$ about $q$ cannot contain points from $U$. This means that $V$ is open.

Then $U$ and $V$ form a separation of $S$, which is a contradiction so $V$ must be empty since $U$ is not.

*Remark: all we used here is convexity of balls in $\mathbb R^n$. The more fundamental condition, is local path-connectedness. We have: locally path-connected+connected = path-connected.


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