Let $X= \mathbb{R} \times \{0,1\}$ with the usual topology. Under the relation $R$, $(a,b)=(c,d)$ whenever $a=c<0$ and every other point is identified only with itself. Form the quotient space $Y= (\mathbb{R} \times \{0,1\} ) / R$.
I want to show that the Y is connected.
Here are my thoughts:
I think two approaches might work: showing that Y is path-connected or showing that Y can’t be disconnected ( via argument by contradiction).
I try the second approach: If $\{A,B\}$ is an open partition of $Y$ then wlog $A$ contains $[\{(0,0)\}]$. Then $[(B(0,r)\times \{0\}) \cup (B(0,r’)\times \{1\})] \subset A$. Hence B can’t be open ( or $B=\emptyset) $because it has “endpoints” $\{(r,0)\}$ and $\{(r,1)\}$. We get the desired contradiction.
Is this correct? Many thanks!
I don't quite follow your argument. Nevertheless note that the only partition of $\mathbb{R}\times\{0,1\}$ into two nonempty open sets would be given by $\mathbb{R}\times\{0\}$ and $\mathbb{R}\times\{1\}$. Let $f: \mathbb{R}\times\{0,1\} \rightarrow \mathbb{R}\times\{0,1\}/R$ be the quotient map. If $\mathbb{R}\times\{0,1\}/R$ can be partitioned into nonempty open sets $A$ and $B$ then $f^{-1}(A)$ and $f^{-1}(B)$ are nonempty open sets that partition $\mathbb{R}\times\{0,1\}$. So without loss of generality you can assume that $f^{-1}(A)=\mathbb{R}\times\{0\}$ and $f^{-1}(B)=\mathbb{R}\times\{1\}$. But this is impossible by definition of $R$.