I wanted to equivalence between both statement

Open nonempty Path connected $\implies$ Connected
Open nonempty Connected $\implies$ Path connected

My attempt:
Let $\sigma$ be path connected set .i.e for any $x,y\in \sigma$ there exist curve joing these point inside $\sigma$.
On contrary assume it is not connected set .So
$\sigma=\sigma_1\cup \sigma_2$ where $\sigma_1$ and $\sigma_2$ is disjoint open set and $w_1\in \sigma1$ and $w_2\in \sigma2$.
Let $\gamma$ be curve joining $w_1,w_2$ Define $Z:[0,1]\to \sigma$ such that $Z(0)=W_1$,$Z(1)=w_2$
Let $t^*$=sup{$t|z(t)\in \sigma$}
Now .
Case 1:$z(t^*)\in \sigma_1$
As $ \sigma_1$ is open $\exists r>0$ such that $B(z(t^*),r)\subset \sigma_1$ So there exist $t_1>t^*$ such that $z(t_1)\in \sigma_1$ Please check this
which is contradicts with sup assumption.
Case 2:$z(t^*)\in \sigma_2$
As $ \sigma_2$ is open $\exists r>0$ such that $B(z(t^*),r)\subset \sigma_2$ So there exist $t_2<t^*$ such that $z(t_2)\in \sigma_1$
Which is contradiction.

So $t^*\notin \sigma$ as $\sigma=\sigma_1\cup \sigma_2$.

Second part:
Let $\sigma$ is open and connected set .
Assume it is not path connected.$x\in \sigma$
$\sigma_1$={$y\in \sigma$|x is connected to y by curve}
$\sigma_2$={$y\in \sigma$|x is not connected to y by any curve}
$x\in \sigma_1 \implies \sigma_1\neq \phi$
Claim : $\sigma_1 ,\sigma_2$ are open and disjoint set.
$\sigma=\sigma_1 \cup \sigma_2$ but $\sigma$ is open
$z\in \sigma_1 $ so there exist curve form x to z .
$\sigma$ is open then $\exists r>0$ in $B(z,r)\subset \sigma$
That means in that nbhd point also are connected by curve from z .As every open ball is connected .Please check this argument
Hence $B(z,r)\subset \sigma_1 $ hence z is interior point. This is true for every point in $\sigma_1$ Hence $\sigma_1$ is open.
If $\sigma_2$ is nonempty then it will contradicts with fact that $\sigma =\sigma_1 \cup \sigma_2$ is connected.
Hence it is path connected.
Any Help will be appreciated

  • $\begingroup$ Which topological space is this happening in? $\endgroup$ – Jack M Aug 19 '18 at 11:36
  • $\begingroup$ Sir I don't have learned any topology yet.<br/>It is complex plane $\endgroup$ – SRJ Aug 19 '18 at 11:41
  • $\begingroup$ Your first proof may be more elaborate than it needs to be. Since $Z$ is continuous, $Z^{-1}(\sigma_1)$ and $Z^{-1}(\sigma_2)$ are open, implying that $[0,1]$ is disconnected. If you know that $[0,1]$ is connected, yo're done. $\endgroup$ – bof Aug 19 '18 at 11:49
  • $\begingroup$ For the second proof, trying to prove the set is open by showing that the complement is closed is a bad plan, Prove openness directly! Find a disk $D$ about $z$ that's contained in $\sigma.$ If you can get from $x$ to $z$ (via a continuous path in $\sigma$) then you can get from $x$ to any point $w$ in $D.$ First wend your way from $x$ to $z,$ then go in a straight line from $z$ to $w.$ Disks are convex! $\endgroup$ – bof Aug 19 '18 at 11:58
  • $\begingroup$ @bof Sir I think I had done same .<br/> Please have a look.Actually initially I thought when posting on that way.But after some time I had got idea and I had edited answer.Thanking You $\endgroup$ – SRJ Aug 19 '18 at 12:00

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