Nested homeomorphic sets

Question

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!

Answer 1

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.

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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 $\bigcup_{i \geq 1} U_i = S^1$.

Answer 2

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) $.