Homeomorphisms between two spaces with different topologies

Question

As far as I know, the main goal of general topology is to find relations between "different looking" objects which have the same structure and thus share interesting properties, such as (path-)connectedness, compactness etc.

The trivial and the discrete topologies, say $\tau_T, \tau_D$, are two of the first examples of topologies on a set $X$, as $X, \emptyset \in \tau_T$ and $X, \emptyset \in \tau_D$, any arbitrary union of open sets is open and the intersection of any finite number of open sets is open. This led me to the following question:

Do we study these topologies just because they are "obvious" collections of subsets of $X$ which satisfy the three conditions of the definition?

Firstly, I had started to wonder if any non-trivial homeomorphism between a discrete and a space with a less simple topology existed.

MY THOUGHTS: Soon I came up with this fact: if $(X, \tau_D)$ is a discrete space, $(Y, \tau)$ is any space and $f: X \to Y$ is a homeomorphism, then $Y$ also has the discrete topology, since points in $Y$ are also open due to the fact that $f$ is both surjective and open.

Moreover, I know that a continuous function from a trivial space to a $T_0$ space is constant as shown here, so it is not bijective if the domain has more than one element; hence, there is no non-trivial homeomorphism between a space with one of these topologies and another space with a less simple topology. Therefore, here comes my question:

Do homeomorphisms between sets with different topologies actually exist?

I hope so, as it would be extremely interesting for me. I would really appreciate as many examples as possible of these homeomorphisms!

NOTE: I know some homeomorphisms involving quotient spaces, like $S^1$ being homeomorphic both to the projective line $\mathbb{P^1(R)}$ and to $\mathbb{R/Z}$.

However, I find these examples involving quotient topologies less interesting, as by definition they depend on other topologies (the quotient topology is a coinduced topology - it says it all). I would like to know some new homeomorphisms between spaces having independent topologies, for example $\mathbb{R}$ with the Sorgenfrey line topology and another set with its own topology.

Answer 1

Per OP’s request, here is an example in my comments, promoted to an answer:

$\mathbb{R}$ equipped with the lower limit topology (i.e., the topology generated by $[a, \infty)$ and $(-\infty, b)$, $a, b \in \mathbb{R}$, as a subbasis), aka the Sorgenfrey line, is homeomorphic to $\mathbb{R} \times \{1\} \subset \mathbb{R} \times \{0, 1\}$ equipped with the subspace topology, and where the latter space is equipped with the order topology obtained from the lexicographic ordering. The homeomorphism simply sends $x \in \mathbb{R}$ to $(x, 1) \in \mathbb{R} \times \{1\}$.

Recall that, if $(X, <)$ is a (partial) order, then the order topology on $X$ is the topology generated by $\{x \in X: x > a\}$ and $\{x \in X: x < b\}$, $a, b \in X$, as a subbasis. Recall that, if $(X, <_X)$ and $(Y, <_Y)$ are (partial) orders, then the lexicographic ordering $<$ on $X \times Y$ is given by,

$$(x_1, y_1) < (x_2, y_2) \Leftrightarrow x_1 < x_2 \text{ or } (x_1 = x_2 \text{ and } y_1 < y_2)$$

Thus, the subspace topology on $\mathbb{R} \times \{1\}$ is generated by the following four kinds of sets,

$$\{(x, 1): (x, 1) > (a, 1)\} = (a, \infty) \times \{1\}$$

$$\{(x, 1): (x, 1) > (a, 0)\} = [a, \infty) \times \{1\}$$

$$\{(x, 1): (x, 1) < (b, 1)\} = (-\infty, b) \times \{1\}$$

$$\{(x, 1): (x, 1) < (b, 0)\} = (-\infty, b) \times \{1\}$$

Since $(a, \infty)$ is open in the lower limit topology, from this description it is easy to check that $\mathbb{R} \ni x \mapsto (x, 1) \in \mathbb{R} \times \{1\}$ is a homeomorphism, as claimed.

As for $\mathbb{Z} \times [0, \infty) \times \{1\} \subset \mathbb{Z} \times [0, \infty) \times \{0, 1\}$, this just follows from $\mathbb{Z} \times [0, \infty)$ with lexicographic ordering is order-isomorphic to $\mathbb{R}$ - first choose an order-isomorphism from $[0, \infty)$ to $[0, 1)$, then sends $(z, r) \in \mathbb{Z} \times [0, 1)$ to $z + r \in \mathbb{R}$. The result now follows the corresponding result on $\mathbb{R} \times \{1\} \subset \mathbb{R} \times \{0, 1\}$.