Preamble
As we have seen $(0,1)\cup(1,2)$ is not a connected subset of $\mathbb R.$ However both (0,1) and (1,2) are connected, and further more, for each point in (0,1) ,(0,1) is the largest connected subset of $(0,1) \cup (1,2)$ containing the point. Similarly,(1,2) is the largest connected subset of $(0,1)\cup(1,2)$ containing each of its points
The same “amount of of dis-connectivity” does not guarantee homeomorphism.
Problem Statement
Let $X=(0,1) \cup (1,2)$ with the relative topology inherited from the usual topology on R and let Y={a,b} with the discrete topology. Show that there is a 1-1 correspondence between the components of X and the components of Y,but X and Y are not homeomorphic
What I know
(0,1),(1,2) are intervals and infinite
In general if a$\ne$ b (a,b)~R
Y is discrete and finite. No bijection between X and Y ,thus no Homeomorphism.
Components of X are (0,1),(1,2) with two end points per interval.
Attempted proof
0$\leftrightarrow$ {a}
1$\leftrightarrow$ {b}
It’s bijective.
This is probably wrong.
I could use the S-B theorem
Am l on the correct path?
Hints: $X = (0, 1) \cup (1, 2)$ has two connected components and so you can find a $1$-$1$ correspondence (or bijection if you prefer) between the set of connected components of $X$ and any set with two elements. If you give any set $Y$ the discrete topology, then its connected components are the singleton sets $\{y\}$ for $y \in Y$. So if $Y = \{a, b\}$ (where $a\neq b$) is given the discrete topology it has two connected components: so its connected components can be put in $1$-$1$ correspondence with those of $X$. If $X$ and $Y$ were homeomorphic, they would have the same number of elements, but $X$ is infinite and $Y$ is finite.