2
$\begingroup$

Question:

Show that $\left ( 0,1 \right )$ and $\mathbb{R}$ are homeomorphic.

Proposition:If $a,b,c,d \in \mathbb{R}$, $a<b$ and $c<d$ then $\left ( a,b \right )$ is homeomorphic to $\left ( c,d \right )$.

For any point $x \in \mathbb{R}$, we have that $x \in \left ( c,d \right )$ where $c<d$

Hence, $\left ( 0,1 \right )$ is homeomorphic to $\mathbb{R}$

Now,

Question: Show that $\left ( 0,1 \right ) \cup \left ( 2,3 \right )$ is not homeomorphic to $\mathbb{R}$.

$\textbf{Hint}$: any two points in $\mathbb{R}$ are joined by a continuous curve.

I'll like a hint as to the second question.

Any help is appreciated.

Thanks in advance.

$\endgroup$
  • $\begingroup$ For $x\in (0,1)$ let $f(x)=\tan \pi (x-1/2).$ Then $f:(0,1)\to \mathbb R$ is a homeomorphism. This is because $f$ is a continuous, strictly monotonic bijection. $\endgroup$ – DanielWainfleet Sep 22 '16 at 1:14
1
$\begingroup$

Suppose you have a homeomorphism $\phi: (0,1) \cup (2,3) \to \mathbb{R}$. Take the points $x=\phi(0.5)$ and $y=\phi(2,3)$. Now have a continuous curve from $x$ to $y$, lets call it $\gamma: [0,1] \to \mathbb{R}$ (I think you can construct such a curve easily).

Hint: The pullback of the curve $\gamma'=\phi^{-1}(\gamma)$ is hence also a continuous curve in $(0,1) \cup (2,3)$ with certain endpoints. Does such a curve exist?

Another possibility/hint without curves: Make yourself clear that in $(0,1) \cup (2,3)$ the interval $(0,1)$ is open (as open interval) and closed (as complement of the open interval $(2,3)$). Assume you have a homeomorphism $\phi: (0,1) \cup (2,3) \to \mathbb{R}$, then $\phi( (0,1)) \subseteq \mathbb{R}$ is (since $\phi$ is a homeomorphism) a closed and open subset of $\mathbb{R}$. You might know all the subsets of $\mathbb{R}$ which are closed and open (hint: there are 2 such subsets). Take it from here (remember that images of non-empty sets are non-empty and a homeomorphism is also injective)!

$\endgroup$
  • $\begingroup$ So essentially what I do need to do is find an inverse function such that for some value in the codomain, the inverse function maps that value to a value not in the domain? $\endgroup$ – Mathematicing Sep 21 '16 at 13:52
  • $\begingroup$ @Mathematicing Nope. I suppose you follow the first hint: You get that inverse function (since you have an homeomorphism). What you should focus on is the path $\gamma '$. This is a continuous path from $\phi^{-1}(x)=0.5$ to $\phi^{-1}(y)=1.5$ in your union of intervals. You now have to show, that such a path is not possible (if you didn't see that yet). Hint: Look at the point, where you "jump" from one intervall to the next one in the domain of your path. (actually I prefer the way without curves but maybe the first way is easier to see for a start) $\endgroup$ – ctst Sep 21 '16 at 13:58
  • $\begingroup$ I prefer the hint without curves. I assume the function is a homeomorphism. So we have that the function f is bijective with f and f^{-1} being continuous. We know that open intervals are open sets. The union of open sets are open sets. By this, we have that f and f^{-1} maps open sets to open sets. Have I got this correct? $\endgroup$ – Mathematicing Sep 21 '16 at 14:06
  • $\begingroup$ @Mathematicing Yes absolutly (you don't need that the union of open sets are open sets for this, but only that $f$ and $f^{-1}$ are continuous)! And by taking complements you also know $f$ maps closed sets to closed sets. $\endgroup$ – ctst Sep 21 '16 at 14:08
  • $\begingroup$ Would you mind expounding on the part where you mention f maps closed sets to closed sets? I think there's a subtlety that I'm not realising it fully. $\endgroup$ – Mathematicing Sep 21 '16 at 14:17
3
$\begingroup$

Let's go back to your first problem:

For any point $x \in \mathbb{R}$, we have that $x \in \left ( c,d \right )$ where $c<d$

Hence, $\left ( 0,1 \right )$ is homeomorphic to $\mathbb{R}$

One point that's confusing for a reader is that you say "hence" as if the second sentence follows from the first. But the two don't seem to have much to do with each other.

The first statement looks like you've jumbled up the quantifiers. You're asked to prove that for all $a$, $b$, $c$, and $d$ with $a<b$ and $c<d$, that $(a,b)$ is homeomorphic to $(c,d)$. You assert that for all $x \in \mathbb{R}$, that there exist $c$ and $d$ such that $c<x<d$. This is true, but it's not relevant to the problem.

First, try to show that $(0,1)$ is homeomorphic to $\mathbb{R}$. Then, generalize your proof to show that $(a,b)$ is homeomorphic to $\mathbb{R}$, for any $a<b$. This shows that $(a,b)$ is homeomorphic to $(c,d)$, by transitivity.

To look for a continuous, invertible map that sends $(0,1)$ to $(-\infty,\infty)$, think of rational functions. Can you construct a function $f(x)$ which is monotone increasing on $(0,1)$, has $\lim_{x\to 0^+} f(x) =-\infty$, and $\lim_{x\to 1^-}f(x) = +\infty$?

$\endgroup$
1
$\begingroup$

I am not exactly sure what you did to prove the first part... But regarding your second question, $\mathbb{R}$ is connected, whereas $(0,1) \cup (2,3)$ is not, and homeomorphisms preserve connectedness.

$\endgroup$
  • $\begingroup$ In the first part, all I note was that any value on the real line is between two open intervals. Then, using the proposition that the open intervals is an interval between c and d with c <d. $\endgroup$ – Mathematicing Sep 21 '16 at 12:58

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.