# Whether the subspaces are homeomorphic or not

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.

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 $$Y$$.

• By connected components you mean maximal connected subset of the topological space? Commented Feb 12, 2021 at 9:23
• Yes, that's what “connected component” means. Commented Feb 12, 2021 at 9:24
• In the first paragraph you have written "But it has four instead" do you mean "But it has two instead " Commented Feb 12, 2021 at 10:49
• No. It turns out that, when I wrote “four”, I actually meant “four”. Commented Feb 12, 2021 at 10:52
• Right. I thought that you were talking about “since $Y\setminus\{(-1,0),(3,0),(1,0)\}$ has four connected components”. My bad. I've edited my answer. Commented Feb 13, 2021 at 11:17

$$X$$ 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.