Whether the subspaces are homeomorphic or not

Question

From topology without tears,

Let $X$ be a unit circle in $\mathbb{R}^2$,that is , $X=\{(x,y):x^2+y^2=1\}$ and has subspace topology. $Y$ be a subspace in $\mathbb{R}^2$ given by $Y=\{(x,y):x^2+y^2=1\} \cup \{(x,y):(x-2)^2+y^2=1\} $ and $Z$ be a subspace in $\mathbb{R}^2$ given by $Z=\{(x,y):x^2+y^2=1\} \cup \left\{(x,y):\left(x-\frac{3}{2}\right)^2+y^2=1\right\} $

I have two questions:

1)How to show whether $Y$ is homeomorphic to a interval?

I have theorem:Let $f:(X, \tau_1) \rightarrow (Y, \tau_2)$ be a homeomorphism. Let $a \in X$, so that $X \setminus \lbrace a \rbrace$ is a subspace of $X$ and has induced topology $\tau_3$. Also, $Y\setminus\lbrace f(a)\rbrace$ is subspace of $Y$ and has induced topology $\tau_4$. Then $(X \setminus \lbrace a \rbrace, \tau_3)$ is homeomorphic to $(Y\setminus\lbrace f(a)\rbrace, \tau_4)$.

Using the above theorem I can show $X$ is not homeomorphic to any interval say $(a,b)$.Because if it is,then $X\setminus\{a\}$ must be homeomorphic to $(a,b)\setminus\{f(a)\}$ for all $a \in X$ by above theorem. But $X\setminus\{a\}$ is always connected but $(a,b)\setminus\{f(a)\}$ is not connected. Therefore $X$ cannot be homeomorphic to any interval.

But how should I show whether $Y$ is homeomorphic to any interval or not.Because here $Y\setminus\{(1,0)\}$ is not connected.So I cannot use the same argument as above and show that $Y$ is not homeomorphic to any interval.Is it homeomorphic?How should I show that?

2)To show that $Z$ is not homeomorphic to $X$ and $Y$

Using the above theorem I can show that $Z$ is not homeomorphic to $Y$,Since $Z-\{a\}$ is always connected but $Y\setminus\{(1,0)\}$ is not connected so I can define $f(a)=(1,0)$ and show $Z\setminus\{a\}$ cannot be homeomorphic to $Y\setminus\{f(a)\}$ for some $a$. And hence $Z$ cannot be homeomorphic to $X$

But how should I show $Z$ is not homeomorphic to $X$,because $Z\setminus\{a\}$ and $X\setminus\{b\}$ is always connected for all $a\in X$ and $b\in Z$. Connectedness cannot be used here.So how should I prove this?

Any hint will be of big help.

Thanks in advance!

Answer 1

Since $Y$ is compact, if it is homeomorphic to some interval, then it has to be an interval of the form $[a,b]$. Suppose that there is such a homeomorphism $f\colon Y\longrightarrow[a,b]$. Note that $Y\setminus\{(-1,0),(3,0)\}$ is connected. But the only two points $c$ and $d$ from $[a,b]$ such that $[a,b]\setminus\{c,d\}$ is connected are $a$ and $b$. So, $f(-1,0)=a$ and $f(3,0)=b$ or $f(-1,0)=b$ and $f(3,0)=a$. Now, let $p=f(1,0)$. Then, since $Y\setminus\{(-1,0),(3,0),(1,0)\}$ has four connected components, $(a,b)\setminus\{p\}$ should also have four connected components. But it has two instead.

And there are two points $x$ and $y$ in $Z$ such that $Z\setminus\{x,y\}$ has four connected components. No such points exist in $X$ or in $Y$.

Answer 2

$X$ has no cutpoints at all

$Y$ has a unique cutpoint.

$Z$ has the property that no removal of two points simultaneously makes $Z$ disconnected. (And it has no cutpoints).

This topologically distinguishes these spaces, as cutsets and cutpoints are preserved by homeomorphisms.

Note that intervals have at most two non-cutpoints.