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Suppose we have a countable collection of sets $\{U_n\}$ such that $U_n\subset U_{n+1}$ for each $n$ and $U_n$ is homeomorphic to $\mathbb{R}$ (or more generally, $X$) for each $n$, then is $\bigcup_{n=1}^\infty U_n$ homeomorphic to $\mathbb{R}$ (or $X$)? I am not sure pushing $n$ to $\infty$ works and I couldn't construct an explicit homeomorphism. Thanks!

bof
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J.Doe
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2 Answers2

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Let $$U_i = \{e^{2\pi i \theta} -1 \mid \theta \in [0,1-1/i)\} \cup \{e^{2\pi i \theta} +1 \mid \theta \in [1/2, 3/2-1/i)\}.$$

Then every $U_i$ is a subset of the union of two circles that are glued together at a single point at the origin and is homeomorphic to $\mathbb{R}$. The union $\bigcup_i \geq 1$ is equal to the union of the two circles.

Here's a simpler example for $X\cong [0,1)$ using the same idea. Let $$U_i = \{e^{2\pi i\theta} \mid \theta \in [0,1-1/i)\}.$$

Then $U_i \cong [0,1)$ for all $i \geq 1$, but $\bigcup_{i \geq 1} U_i = S^1$.

Dan Rust
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    I think this would be a better example if you simply gave your solution to the $\mathbb R$ case (without the "wedge" language, which people may not understand). – zhw. Jan 26 '19 at 02:15
  • @zhw. thanks for the suggestion. Hopefully the edit is a bit easier to read (unfortunately, it's not too easy to write the $U_i$s in a very simple way without talking about gluing/wedges). – Dan Rust Jan 26 '19 at 02:23
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    Oh, right. Or $U_n={t:-2\lt t\le1}\cup{e^{it}:0\le t\lt\pi-\frac1n}$. By the way, you are using the letter $i$ for two different things. – bof Jan 26 '19 at 02:47
  • Yes, I think the edit is good. +1 – zhw. Jan 26 '19 at 06:33
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For $X=\Bbb R$, it's homeomorphic with any open interval, and $U_n\subseteq U_{n+1}$ implies here that $U_n$ is a subinterval.
Thus, we can choose decreasing $a_n$ and increasing $b_n$ such that $U_n$ is homeomorphically mapped to $(a_n, b_n)$.
Then, $\cup_nU_n$ will be homeomorphic with $(\inf a_n, \sup b_n) $.

Berci
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