There are 2 topological spaces:
$\mathcal{T}_1$ consists of $\mathbb{R}$, $\emptyset$, and every interval $(−n,n)$, for n any positive integer;
$\mathcal{T}_2$ consists of $\mathbb{R}$, $\emptyset$, and every interval $[−n,n]$, for n any positive integer;
I was thinking that the spaces were not homeomorphic but than I constructed a homeomorphism using induction:
- For n = 1. The intervals are homeomorphic $[-1, 1] \cong (-1, 1)$
$$ f: [-1, 1] \rightarrow (-1, 1)\\ f(x) = \begin{cases} -\frac{1}{2} & \text{for } x = -1 \\ \frac{1}{2} & \text{for } x = 1 \\ -\frac{x}{2} & \text{for } x = -\frac{1}{2^n}\\ \frac{x}{2} & \text{for } x = \frac{1}{2^n}\\ x & \text{otherwise} \end{cases} $$
- Given that $(-n, n) \cong [-n, n]$ we can prove that $(-(n+1), n+1) \cong [-(n+1), n+1]$. We need to find bijection for the tails. Bijection for the left tail $(-(n+1), -n] \cong [-(n+1), -n)$
$$ f(x) = -x - 2n - 1 $$
Similar for the right tail.
Is my reasoning correct?
Yes, your construction looks OK.
More generally, let $f$ be any bijection from $\mathbb{R}$ to $\mathbb{R}$ such that $f([-n,n])=(-n,n)$, for all positive integers $n$, and let $g=f^{-1}$.
Then $f$ is continuous, as a function from $(\mathbb{R},\tau_2)$ to $(\mathbb{R},\tau_1)$, since for all positive integers $n$, $$f^{-1}((-n,n))=[-n,n]$$ and $g$ is continuous, as a function from $(\mathbb{R},\tau_1)$ to $(\mathbb{R},\tau_2)$, since for all positive integers $n$, $$g^{-1}([-n,n])=(-n,n)$$