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Consider the following subspaces of $\mathbb{R}$ with the usual topology: $$X = (0, 1) \cup \{2\} \cup (3, 4) \cup \{5\} \cup \cdots \cup (3n, 3n + 1) \cup \{3n + 2\} \cup\cdots$$ $$Y = (0, 1] \cup (3, 4) \cup \{5\} \cup\cdots\cup (3n, 3n + 1) \cup \{3n + 2\} \cup\cdots$$ Is $X$ homeomorphic to $Y$ ?

For $X$ to be homeomorphic to $Y$, we need to specify a bijective function from $X$ to $Y$ and inverse function from $Y$ to $X$ are continuous. From $(3,4)$ onwards, we can map by identity function. How can I map $(0,1) \cup \{2\}$ to $Y$? $(0,1]$, in usual topology is not open and closed. Can I write $(0,1]$ as $(0,1)\cup\{1\}$, and then map $\{0,1\}$ by identity map and $\{1\}$ to $\{2\}$. Please forgive me if any of what I think is stupid.

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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Mar 3 '13 at 6:32

Hint: If $f:A\to B$ is a continuous map between topological spaces, and $R$ is a connected component of $A$, then there is a connected component $S$ of $B$ such that $f(R)\subseteq S$.

What does this imply about how homeomorphisms map the connected components of spaces?

Do you see how to apply this to your situation?

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Each of the sets in X and Y are connected,except for(0,1],in Y.So(0,1] can be writen as,(0,1)U{1},and (0,1) maps to (0,1) under identity map,and i can map {2} to {1}? – Ram Tvm Mar 3 '13 at 7:56
That's not correct; the set $(0,1]$ is connected. – Zev Chonoles Mar 3 '13 at 8:04
It all comes down to whether there is a homoemorphism between $(0,1)$ and $(0,1]$. The singletons can be handled by mapping $\{k\}$ from $X$ to $\{3+k\}$ from $Y$ – Ross Millikan Mar 3 '13 at 15:21


  1. If $f:X\to Y$ is a homeomorphism, then the disconnected singletons of $X$ must map to the disconnected singletons $Y$. Reason being that they are isolated points (open subsets) in $X$ and thus their images are also isolated points in $Y$.

  2. If $f:X\to Y$ is a homeomorphism, then the image of a connected set is connected. Hence any interval in $X$ must map to an interval in $Y$.

  3. Conclude with 1. & 2. that if $f:X\to Y$ is a homeomorphism, then an interval of the form $(3n,3n+1)$ is homeomorphic to $(0,1]$ for some $n\in\mathbb{N}$. Take the co-restriction of this homeomorphism to the set $(0,1)$, and try to conclude a contradiction with a connectedness argument.

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