I want to show that $(0,1)$ is homeomorphic to $\mathbb{R}$ by finding a homeomorphism between the two. I think the function will be related to $tan(x)$ but I'm stuck on how to modify it to fit the domain $(0,1)$.
Any help would be appreciated!
[Math] Homeomorphism from $(0,1)$ to $\mathbb{R}$
analysismetric-spaces
Related Solutions
Let me outline a proof; I shall leave the details of this proof as exercises.
Exercise 1: Prove that the unit circle $S^1$ is homeomorphic to the space obtained from the unit interval $[0,1]$ by identifying the endpoints $0$ and $1$. (Hint: the exponential mapping $\theta\to e^{i\theta}$ can be composed with a linear mapping to give a homeomorphism.)
Let $C$ be the image of a simple closed curve in the plane, that is, let $C$ be the image of a continuous function $f:[0,1]\to \mathbb{R}^2$ such that $f:(0,1)\to\mathbb{R}^2$ is injective and $f(0)=f(1)$.
Exercise 2: Prove that $C$ is homeomorphic to the unit circle $S^1$. (Hint: show that $f:[0,1]\to C$ induces a continuous bijection $\tilde{f}:S^1\to C$ using Exercise 1 and then show that $f$ is an open mapping using the compactness of $S^1$ and the fact that the plane is Hausdorff.)
We now need an elementary lemma of point-set topology:
Exercise 3: Let $X$ be a topological space and let $X=A\cup B$ where $A$ and $B$ are closed subspaces of $X$. Prove that if $g:A\to Y$ and $h:B\to Y$ are continuous functions into a topological space $Y$ such that $g(x)=h(x)$ for all $x\in A\cap B$, then there is a unique continuous function $f:X\to Y$ such that $f(a)=g(a)$ and $f(b)=h(b)$ for all $a\in A$ and $b\in B$.
Finally, we can prove the result of your question:
Exercise 4: Prove that the unit square is the image of a simple closed curve in the plane and conclude that it is homeomorphic to the unit circle. (Hint: you can use Exercise 3 to "glue" continuous mappings together.)
Exercise 5: Give examples of images of simple closed curves in the plane (using Exercise 3) and conclude (using Exercise 2) that they are homeomorphic to the unit circle $S^1$.
I hope this helps!
Think about how the plane minus a point looks: it is a disjoint union of concentric circles, each circle being of some radius $r \in (0,\infty)$. (This is how polar coordinates work.)
This essentially is the same as saying that $\mathbb R^2 \setminus \{(0,0)\} \cong S^1 \times (0,\infty).$ Now you just have to remember that $(0,\infty) \cong \mathbb R$.
Rather than asking about general methods of finding homeomorphisms (which is a very difficult problem when stated in full generally), based on what you have written, I think that perhaps you should practice visualizing different examples of products. E.g. you might want to convince yourself that a torus (the surface of a donut) is homeomorphic to $S^1 \times S^1$, by thinking of it as a circle of cirles (just as $\mathbb R^2\setminus \{(0,0)\}$ is a circle of intervals). You could also consider the complement of a ball of radius $1$ in the ball of radius $2$, and convince yourself that it is homeomorphic to the product of $S^2$ and a closed interval.
One important "background" fact to keep in mind is that all open intervals are homeomorphic, and that all closed intervals are homeomorphic, so that the particular endpoints don't matter. (In different concrete situations, it can be helpful to think in terms of different endpoints. E.g. $\mathbb R = (-\infty,\infty)$, but in my explanation of your problem above, it was more convenient visually to think in terms of the open interval $(0,\infty)$.)
Best Answer
$\tan:(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb{R}$ is a homeomorphism between $(-\pi/2,\pi/2)$ and $\mathbb{R}$. Define $f:(0,1)\to(-\pi/2,\pi/2)$ by $f(t)=-(1-t)\frac{\pi}{2}+t\frac{\pi}{2}=-\frac{\pi}{2}+t\pi$. Then, $f$ is a homemorphism between $(0,1) $ and $(-\pi/2,\pi/2)$. Therefore, $h:(0,1)\to\mathbb{R}$ given by $h(t)=\tan(f(t))=\tan(-\frac{\pi}{2}+\pi t)$ works.