# Homeomorphisms between two spaces with different topologies

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.

• math.stackexchange.com/q/663231/269764 Commented Sep 2 at 7:10
• What does “sets with different topologies” mean? If two spaces are homeomorphic, I would just say they have “the same topologies”. And about your edit, “spaces having topologies which do not come from a different one”, what exactly does that mean? Like, you just don’t want one of the spaces being constructed as a quotient space? But then why is $P^1(\mathbb{R}) \simeq S^1$ not satisfactory? Commented Sep 2 at 7:35
• @DavideMasi I don’t think there’re many “natural” topologies that are homeomorphic to the Sorgenfrey line, apart from maybe the upper limit topology on $\mathbb{R}$, but I struggle to call that “at first glance completely different”. Though you can certainly create some artificial examples. Say, $\mathbb{R}\times\{1\}\subset\mathbb{R}\times\{0,1\}$ with the subspace topology, where the latter space is equipped with the order topology from lexicographic ordering. If you want it to look even “more different”, you can choose… Commented Sep 2 at 8:53
• @DavideMasi I mean, I suppose from an intuitive point of view, I get what you are trying to say, but it is a bit too subjective. After all, as long as the underlying sets are different, the topologies, as sets, can never be the same. You can, for example, just rename a point (or a few points), then you’ll get a homeomorphic space with different topologies in your sense. This is precisely what happened with the example in the link. In a sense, this is what happens for any pair of homeomorphic spaces. Commented Sep 2 at 8:58
• Please learn how to parse your questions. E.g. Most of your post looks like a ramble/word salad. Perhaps use more organization (paragraphs, perhaps). Commented Sep 2 at 16:12

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 $$\mathbb{R} \times \{1\} \subset \mathbb{R} \times \{0, 1\}$$.

• One can make this even funnier: for $\mathbb{R} \times \{0, 1, 2\}$, the subspace $\mathbb{R} \times \{0\}$ has upper limit topology, $\mathbb{R} \times \{1\}$ has discrete topology, and $\mathbb{R} \times \{2\}$ has lower limit topology. Commented Sep 2 at 9:30
• @DavideMasi For example, the one-point compactification of $\mathbb{R}^n$ is homeomorphic to the unit sphere of $\mathbb{R}^{n+1}$ for all $n\geq 0$. The one-point compactification of $\omega_1$ is homeomorphic to $\omega_1+1$. The unit sphere of any infinite-dimensional Hilbert space $H$ (say, $H=\ell^2$) is homeomorphic to $H$. $\ell^\infty$ is homeomorphic (in fact, linearly homeomorphic) to $L^\infty(\mathbb{R})$, where both are equipped with the norm topology (despite them not being linearly isometric). Commented Sep 2 at 10:03
• @DavideMasi The Cantor space (constructed as a subspace of $\mathbb{R}$) is homeomorphic to $\{0,1\}^\mathbb{N}$ equipped with the product topology. It is also homeomorphic to the Stone space of the free Boolean algebra on countably infinite many generators (or any (nontrivial) countable atomless Boolean algebra). The space of irrational numbers is homeomorphic to the Baire space, i.e., $\mathbb{N}^\mathbb{N}$ equipped with the product topology. The Stone-Cech compactification of $\mathbb{N}$ is homeomorphic to the Stone space of the Boolean algebra $P(\mathbb{N})$ and also homeomorphic to… Commented Sep 2 at 10:09
• @DavideMasi … the spectrum of $\ell^\infty$. Similarly, the Stone-Cech compactification of $\mathbb{R}$ is homeomorphic to the spectrum of $C_b(\mathbb{R})$. The primitive ideal spectrum of $B(H)$ is homeomorphic to the Sierpiński space, etc., etc. Commented Sep 2 at 10:13
• @DavideMasi No problem! Some of these examples are relatively easy to see once you know the definitions, but there are a few that are quite specialized and deep results from various different fields. Commented Sep 2 at 10:15