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$ L:= \{(1,0)\}\;\cup\; \bigcup_{n=1}^\infty L_n$ where $L_n:= \{(x,{x\over n})\in \Bbb R^2\mid \; x\in [0,1]\},\; n\in \Bbb N$.


I've been provided that the set L as an example of Connected but not Path-connecte set.

I would like to prove that the given L is connected but not-path-connected?

Below is my proof for connectedness, please check whether it's correct.

Proof of Connected-ness>>

If we choose two open sets $u,v$ in $\Bbb R^2 $ which are $ u = (1,0)$ and $ v = L \setminus \{(1,0)\} $ then A is connected.

Proof of Path-Connected >>

How to do this?

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  • $\begingroup$ The proof for connectedness needs to work for all open sets $u,v$. Also, your $u$ is not open. $\endgroup$ – Hagen von Eitzen May 8 '17 at 10:52
  • $\begingroup$ How could I make it general for all arbitral cases? $\endgroup$ – Beverlie May 8 '17 at 11:07
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Proof of connectedness

Let $U,V\subseteq \Bbb R^2$ be arbitrary open sets with $L\subseteq U\cup V$ and $L\cap U\cap V=\emptyset$. We want to show that $L\subseteq U$ or $L\subseteq V$.

One of them, wlog. $U$, contains the point $(1,0)$, hence also an open disk $B_r(1,0)$ of some positive radius $r$ around that point. For $N>\frac1r$, we have $(1,\frac1N)\in U$, hence $U$ intersects $L_N$. As $L_N$ is connected (it is homeomorphic with $[0,1]$), we conclude that $L_n\subseteq U$. In particular $(0,0)\in U$. But the $U$ intersects $L_n$ for all $n$, hence $L_n\subseteq U$ for all $n$, i.e., $\bigcup_{n=1}^\infty L_n\subseteq U$. As also $(1,0)\in U$, we have $L\subseteq U$.

Disproof of path connectedness

Assume $L$ were path-connected. Then there exists a path from $(0,0)$ to $(1,0)$, i.e., a continuos map $\gamma\colon [0,1]\to L$ with $\gamma(0)=(0,0)$ and $\gamma(1)=(1,0)$. Let $t_0=\sup\{\,t\in[0,1]\mid\gamma(t)=(0,0)\,\}$. By continuity of $\gamma$, also $\gamma(t_0)=(0,0)$. Let $t_1=\inf\{\,t\in[t_0,1]\mid\gamma(t)=(1,0)\,\}$. Again by continuity of $\gamma$, also $\gamma(t_0)=(1,0)$. Then $p:=\gamma((t_0+t_1)/2)$ is $\ne(0,0)$ and $\ne (1,0)$, hence on one (and only one) $L_n$. But there is no path from $p$ to $(1,0)$ that does not pass through $(0,0)$ (to see this, note that $L\setminus\{(0,0)\}$ is not even connected).

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  • $\begingroup$ I'v learned that for the proof of disconnected as follow: $A \in M $ is disconnected if, for two open sets $u, v$ in metric space with: (i) $(A\cap u)\cap(A\cap v) = \emptyset$ (ii) $A\cap u \neq \emptyset$ (iii) $A\cap v \neq \emptyset$ (iv) $(A\cap u)\cup(A\cap v) = A$ How can I logically construe your definition of connectedness from above definition? $\endgroup$ – Beverlie May 8 '17 at 11:20
  • $\begingroup$ What is the definition of homeomorphic? can't get this part : "As $L_N$ is connected (it is homeomorphic with $[0,1]$), we conclude that $L_n\subseteq U$." $\endgroup$ – Beverlie May 8 '17 at 11:39
  • $\begingroup$ @jackerysmith For the first point, you note that a space $A$ is disconnected if (and only if) we can find open $u,v$ such that (i),(ii),(iii) and (iv) hold. Hagen's equivalent formulation is that a space $A$ is connected if and only if whenever we have $u$ and $v$ such that (i) and (iv) hold, then either (ii) or (iii) does not hold (otherwise, they all hold and $A$ is disconnected). Now note that saying that $A\cap u=\emptyset$ or $A\cap v=\emptyset$, in the presence of (iv), is equivalent to saying that $A\cap v= A$ or $A\cap u=A$; i.e., $v\subseteq A$ or $u\subseteq A$. $\endgroup$ – John Gowers May 8 '17 at 17:04
  • $\begingroup$ For your second question, we say that a continuous map $\phi\colon X\to Y$ is a homeomorphism if it has a continuous inverse. If there is a homeomorphism from $X$ to $Y$, then we say that $X$ and $Y$ are homeomorphic. There's an obvious homemorphism from $[0,1]$ to $L_n$ given by $x\mapsto(x,x/n)$. Now we use the fact that if $X$ is connected and $X$ is homeomorphic to $Y$ then $Y$ is connected. You might like to prove this for yourself. $\endgroup$ – John Gowers May 8 '17 at 17:06
  • $\begingroup$ @John Gowers Got it. One more question - how could we know the fact that "the $U$ intersects $L_n$ for all $n$"? $\endgroup$ – Beverlie May 9 '17 at 1:36
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Your proof for connectedness does not make sense. You need to prove that for all pairs $U,V$ of non-empty open subsets of $L$ such that $U\cup V=L$, you have $U\cap V\neq\emptyset$. Instead, you've only mentioned one choice of $U$ and $V$ - and your set $u$ isn't open anyway.

Note that the set $L\setminus\{(0,1)\}$ is path connected (just take a path to the origin and then back out again) and hence connected. Now notice that $(0,1)$ is contained in the closure of $L\setminus\{(0,1)\}$. You can then apply the following result, which is not too difficult to prove:

Theorem Let $X$ be a connected subset of some space $M$ and suppose that $X\subset Y\subset \text{cl}X$. Then $Y$ is connected.

To show that your space is not path connected, you will need to show that the point $(0,1)$ is not connected to the rest of the space by a path. Specifically, let's show that there is no path from $(0,1)$ to $(1,1)$. Indeed, suppose there were such a path $\gamma\colon[0,1]\to L$.

Let $B$ be the open ball of radius $1/2$ about $(0,0)$. $L\setminus B$ is not connected (it is easy to find open sets $U,V$ that disconnect it), so any path from $(0,1)$ to $(1,1)$ must pass through $B$. Let $$ t=\inf\{s\in[0,1]\colon\gamma(s)\in B\} $$

We claim that if $u< t$, then $\gamma(u)=(0,1)$. Indeed, any path from $(0,1)$ to any point in $L\setminus\{(0,1)\}$ must pass through $B$, so if $\gamma(u)\in L\setminus\{(0,1)\}$ then there must be some $v\le u<t$ such that $\gamma(v)\in B$, contradicting the definition of $t$. Since $\{(0,1)\}$ is closed, it follows that $\gamma(t)=(0,1)$.

But we also have $$ t\in\text{cl}\gamma^{-1}(B)\subset\gamma^{-1}(\text{cl}B) $$

and so $\gamma(t)\in\text{cl}(B)$. But this is a contradiction, since $(0,1)$ is not in the closure of $B$.

I think the proof can be tidied up a bit, but that's the idea.

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