# Union of open sets homeomorphic to $\mathbb{R}$

I am trying to prove the following result.

Let $X$ be a topological space. Suppose $\{U_i\}_{i\in\mathbb{N}}$ is a sequence of open sets such that $U_i\subset U_{i+1}\quad\forall\,i\in\mathbb{N}$, $U_i$ homeomorphic to $\mathbb{R}$, and that $X=\bigcup\limits_{i\in\mathbb{N}} U_i$. I have to prove that $X$ is homeomorphic to $\mathbb{R}$.

My idea was trying to extend the homeomorphism $\varphi_1:U_1\longrightarrow (-1,1)$ to $\varphi_2:U_2\longrightarrow (a,b)$, with $(-1,1)\subset (a,b)$ and $\varphi_2(x)=\varphi_1(x)\quad\forall\,x\in V_1$, but I can't seem to do it.

• Each $U_i$ is connected, hence an open interval. Thus $X$ must be an open interval. Can you continue from there? – Vincent Boelens Sep 29 '16 at 12:08

## 2 Answers

I think you are on the right track. Here is how I would do it; suppose you have constructed $\phi_i : U_i \to (a,b)$ (e.g., $i = 1$), and inductively assume that it is increasing. Then fix a homeomorphism $\phi'_{i+1} : U_{i+1} \to (c',d')$ which is also increasing. Then $\phi'_{i+1}(U_i)$ is an open interval in $(c',d')$, let's say $(a',b')$. Then we rescale and translate the interval $(c',d')$ in such a way that $(a',b')$ gets taken to $(a,b)$ -- call this mapping $r$. Then $r \circ \phi'_{i+1}: U_{i+1} \to (c,d)$ is an increasing homeomorphism, which maps $U_i$ to $(a,b)$. Now define $\phi_{i+1}:U_{i+1} \to (c,d)$ by letting it equal $r \circ \phi'_{i+1}$ outside $U_i$, and letting it equal $\phi_i$ on $U_i$. Then use the pasting lemma to show that this is still an isomorphism.

It is a little fidgety, but it works. Note that it is crucial that $a,b,c,d$ are real numbers, and not $-\infty$ or $\infty$ otherwise the argument doesn't work.

• Thanks, I understand the argument now, but what do you mean by $\phi_i,\phi_{i+1}$ increasing? They are defined on $U_i,U_{i+1}$, which could be anything. – user203327 Sep 30 '16 at 9:56
• Oh, yeah, you're right. I wasn't thinking clearly. In that case, apart from a translation and a dilation, $r$ might also need to be a negation. As long as these things match, you know. – Mees de Vries Sep 30 '16 at 15:47

Hint. There exists a homeomorphism $f_1:U_1\to (A_1,B_1 )$ with $A_1<B_1.$ For each $i$ there exists a homeomorphism $f_{i+1}:U_{i+1} \to (A_{i+1},B_{i+1})$ with $A_{i+1}\leq A_i$ and $B_i\leq B_{i+1},$ such that $f_{i+1}|U_i$ (the restriction of $f_{i+1}$ to the domain $U_i$) is equal to $f_i.$

Then $\cup_{i\in N}f_i:\cup_{i\in N}U_i\to \cup_{i\in N}(A_i,B_i)=C$ is a homeomorphism to a non-empty open real interval $C.$ (...$C$ may be unbounded, and possibly $C=\mathbb R.$ )

• I have a doubt, why can you guarantee that there exists $f_{i+1}$ a "extension" of $f_i$? – user203327 Sep 30 '16 at 9:57
• My response was too long for a comment and it's late (for me). Will reply tomorrow. – DanielWainfleet Oct 1 '16 at 5:39