Consider the following topologies on $\mathbb{R}$:
$\tau_1 = \{\mathbb{R}, \emptyset, (-n, n) \forall n\in \mathbb{Z}^+\}$
$\tau_2 = \{\mathbb{R}, \emptyset, [-n, n] \forall n\in \mathbb{Z}^+\}$
$\tau_3 = \{\mathbb{R}, \emptyset, (-r, r) \forall r\in \mathbb{R}^+\}$
$\tau_4 = \{\mathbb{R}, \emptyset, (-r, r)$ and $[-r, r] \forall r\in \mathbb{R}^+\}$
Problem 1: Show $\tau_1 \not\cong \tau_3$
Problem 2: Is $\tau_1 \cong \tau_2$?
Problem 3: Is $\tau_3 \cong \tau_4$?
Note for beginners: Do not assume facts about $\mathbb{R}$ based on the standard Euclidean topology. These are different topologies, and so they may behave differently on otherwise familiar sets-- for example, $(a,b)$ and $[a,b]$ are non-homeomorphic interval types in the Euclidean topology, but you can't use that here. Work from the definition of Homeomorphism.
Citation: S. Morris, "Topology without Tears", problem 4.3.7 i, ii, iii
$|\tau_1| \neq |\tau_3|$ while a homeomorphism induces a bijection of topologies.
Let $f_1$ be a bijection between $(-1,1)$ and $[-1,1]$ (standard set theory).
We can extend this to a bijection $f_2$ between $(-2,2)$ and $[-2,2]$ (both difference sets are also equinumerous).
So continuing this way, we get a sequence of extensions that are bijections between $(-n,n)$ and $[-n,n]$. The union of these functions is then a bijection of the reals that for each $n$ obeys $f[(-n,n)] = [-n,n]$ (and so also $f^{-1}[-n,n]] = (-n,n)$) and so $f$ is both open and continuous. So $\tau_1 \simeq \tau_2$.
In $\tau_4$ the open sets have the property:
$$\exists U,V \in \tau_4: U \subsetneq V \land |V \setminus U| =2\tag{1}$$
While $\tau_3$ does not have the above property as it obeys the property
$$\forall U,V \in \tau_3: U \subsetneq V \implies |V \setminus U| \text{ is infinite}\tag{2}$$
So $\tau_3 \not \simeq \tau_4$:
If $f: (\mathbb{R}, \tau_4) \to (\mathbb{R}, \tau_3)$ would be a homeomorphism, then for $U,V$ as in $(1)$ we'd have $f[U] \subsetneq f[V]$, both would be open and $f[V \setminus U] = f[V] \setminus f[U]$ would contradict $(2)$.