**Problem: The interval $(-1,1)$ $\subset \mathbb{R}$ and $\mathbb{R}$ itself are homeomorphic in their standard topologies.**

As per definition, we call the underlying bijection of a topological equivalence $h: X \to Y$ a homeomorphism. We say $X$ and $Y$ are homeomorphic.

Below is how I showed it. Kindly check and verify my proof. Let me know if there is something which is wrong or needs to be better in terms of style and notation.

Let us define a mapping $f: (-1,1) \to \mathbb{R}$ by $f(x) = \frac{x}{x^{2}-1}$, then we can clearly say that $f$ is well defined. Now, $f(x\_1) = f(x\_2)$. So, that implies that $\frac{x\_1}{x\_1^{2}-1}$ = $\frac{x\_2}{x\_2^{2}-1}$, and solving this you will get $(x\_2-x\_1)(x\_1x\_2 + 1) = 0$ which means either $x\_1 = x\_2$ or $x\_1x\_2 = -1$. Now, if $x\_1x\_2 = -1$, then this condition wont work since $x\_1x\_2 \in (-1,1)$, and therefore its only one condition, $x\_1 = x\_2$. So, $x\_1 = x\_2$ means that $f$ is clearly injective here. Let $y \in \mathbb{R}$ be any element and $f(x) = y$, then $\frac{x}{x^{2}-1} = y$, and solving that gives us $x^2 y - x-y = 0$. Now, $x = \frac{1 + \sqrt(1+ 4y^2)}{2y}$, and $x = \frac{1 - \sqrt(1+ 4y^2)}{2y}$. This is the pre-image of $y$. Since, $x = \frac{1 + \sqrt(1+ 4y^2)}{2y}$ $\notin (-1,1)$ , therefore each element in $\mathbb{R}$ has a pre-image in $(-1,1)$. So, $f$ is surjective. Since, $f$ is injective as well as surjective, therefore its bijective. We now show that $f$ is continuous.

Let us consider a sequence $x\_n$ in $(-1,1)$ that is converging to $x$, then $f(x\_n) = \frac{x\_n}{x\_n^{2}-1}$, and this $\to$ $\frac{x}{x^{2}-1} = f(x)$, since $x\_n \to x$. So, $f(x\_n) \to f(x)$. Now, I have used the concept that if there is any sequence $x\_n \to x$, then => $f(x\_n) \to f(x)$ in the usual topology in $\mathbb{R}$, then $f$ is continuous. Hence, we get that $f:(-1,1) \to \mathbb{R}$ is continuous. Now, lets check its inverse.

$f^{-1}(y)$ = {$f(x) = y | x \in (-1,1)$}, therefore $\frac{x}{x^{2}-1} = y$, where $x \in (-1,1)$, so $x = \frac{1 - \sqrt(1+ 4y^2)}{2y}$, therefore $f^{-1}(y)$ = $\frac{1 - \sqrt(1+ 4y^2)}{2y}$, and by same argument we can say that its inverse is also continuous. 

Thus, we have: 1. $f$ is bijective, 2. $f$ is continuous, and 3. $f^{-1}$ is also continuous. Hence, its a homeomorphism and thus $(-1,1)$ and $\mathbb{R}$ are homeomorphic. 

**Also, one question: Below is the definition of a function to be continuous and open. I actually didnt use that while showing the sequences. Is there a way it can be done using below definition? If yes, please help me on that.
Let $(X, T\_X)$ and $(Y, T\_Y)$ be topological spaces. A function $f: X \to Y$ is said to be continuous iff (if and only if) for every $V \in T\_Y$, $f^{-1}(V) \in T\_X$, and a function $f: X \to Y$ is said to
be open iff (if and only if) for every $U \in T\_X$, $f(U) \in T\_Y$.**

Appreciate your help and support. :)