Are topologies on $\Bbb R$ with bases $\{[-n,n]:n\in\Bbb N\}$ and $\{(-n,n):n\in\Bbb N\}$ homeomorphic?

Question

I think that NO because there is no way to map an open set of the kind $[-n_1,n_1]$ to some open map of the kind $(-n_2,n_2)$.

Proof: imagine some homeomorphism $f:(\Bbb R, T_1) \to (\Bbb R, T_2)$ exist, where $T_1$ have base $B_1=\{[-n,n]:n\in\Bbb N\}$ and $T_2$ base $B_2=\{(-n,n):n\in\Bbb N\}$. Then taking some $f(n)=y$ we need that $y$ will be the max of some $(-n,n)$ (because $n$ is the max of some open set on $T_1$) but by definition open intervals dont have maximum so it is impossible.

Questions:

1. Is my proof and assumptions correct?

2. There is a different way to prove this result (or the opposite if Im wrong)? Can you outline to me someone?

Thank you in advance.

Answer 1

1. You seem to be assuming that the homeomorphism preserves order, or at least that it preserves the maximum or supremum of certain sets, which is not guaranteed for a general homeomorphism.

2. However, even the most general homeomorphism is a bijection so it must preserve the cardinality of sets. This property can be used to establish that the topologies are non-homeomorphic, because one of the two topologies has a nontrivial finite open set while the other one does not.

(adapted following comment)

2. If $n=0$ is not counted among the natural numbers then there exists a homeomorphism, based on the observation that any two intervals of nonzero length have the cardinality of the continuum and are therefore bijective. Choose a bijective mapping $\phi_0:[-1,1]\to(-1,1),$ and for every $n=1,2,\ldots$ choose a bijective mapping $\phi_n:[-n-1,-n)\cup(n,n+1]\to(-n-1,-n]\cup[n,n+1)$. The domains of these mappings form a partition of $\mathbb R;$ and so do the images of these mappings, so they can be glued together to a single bijective mapping $\phi:\mathbb R\to\mathbb R$ that maps the bases of the two topologies, and hence all open sets, to each other.

Answer 2

It is enough to construct a bijection $f: \mathbb{R} \to \mathbb{R}$ such that for all integers $n \ge 1$ you have $f((-n, n))=[-n,n]$.

For $n=1$, we know that there exists a bijection $f_1: (-1,1) \to [-1,1]$ (simply because of a cardinality argument).

For $n \ge 2$ we know that there exists a bijection $$j_{n}: (-n, -n+1] \cup [n-1, n) \to [-n+1, -n) \cup (n-1, n]$$

So we can glue up all the $\{ f_n \}_{n \ge 1}$ to define your $f$. Such a function is bijective and maps $(-n,n)$ to $[-n,n]$ for all $n$.