0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Each $U_i$ is connected, hence an open interval. Thus $X$ must be an open interval. Can you continue from there? $\endgroup$ – Vincent Boelens Sep 29 '16 at 12:08
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user203327 Sep 30 '16 at 9:56
  • $\begingroup$ 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. $\endgroup$ – Mees de Vries Sep 30 '16 at 15:47
1
$\begingroup$

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 $C=\mathbb R.$ )

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.