Are these subsets of $\mathbb{R}$ homeomorphic?

Question

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.

Answer 1

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?

Answer 2

Hints:

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.