Are these two spaces $(\mathbb{R},T_\text{up})$ and $(\mathbb{R},T_\text{low})$ homeomorphic?

Question

This is a topology question, the more formal question is this:

Are the spaces ($\mathbb{R},T_\text{up}$)and ($\mathbb{R},T_\text{low}$) homeomorphic?

Here, $T_\text{up}$ is the upper topology consisting of the sets $\{\emptyset, \mathbb{R}\}\cup\{(a, \infty) \mid a \in \mathbb{R}\}$, and $T_\text{low} := \{\emptyset, \mathbb{R}\} \cup \{(−\infty, a) \mid a \in \mathbb{R}\}$ defines the lower topology on $\mathbb{R}$.

I know these two are homeomorphic if there exists a homeomorphism between them, which is a function $f: (\mathbb{R},T_\text{up}) \to (\mathbb{R},T_\text{low})$ which is bijective, continuous, and has a continuous inverse. However, I'm clueless about how to find such a function or how to prove that they are not homeomorphic.

Any help will be greatly appreciated!

Answer 1

I think that the intuition here is that a negative ray $(-\infty, a)$ and a positive ray $(b,\infty)$ are basically the same thing. Thus the intuitive candidate for a homeomorphism is the map $$ f : (\mathbb{R},T_{\text{up}}) \to (\mathbb{R},T_\text{low}) : x \mapsto -x. $$ I think that it is fairly easy to see that this is a bijection, though here is the proof for completeness:

• Injectivity: Suppose that $x,y\in\mathbb{R}$ with $f(x) = f(y)$. By definition of $f$, this implies that $-x = -y$, which is possible only if $x= y$. Therefore $f$ is injective.
• Surjective: Suppose that $x \in (\mathbb{R},T_\text{low})$. Then $-x \in (\mathbb{R},T_\text{up})$ and $f(-x) = -(-x) = x$. Therefore $f$ is surjective.
• Continuity of $f$: Continuity is slightly more subtle. We want to show that if $V \subseteq \mathbb{R},T_\text{low})$ is open, then $f^{-1}(V) \subseteq (\mathbb{R}, T_\text{up})$ is also open. First, let's deal with the easy cases: if $V= \mathbb{R}$ then $f^{-1}(V) = \mathbb{R}$ (this follows immediately from the bijectivity of $f$), and if $V = \emptyset$ then $f^{-1}(V) = \emptyset$ (this follows from basic theory—there is essentially nothing to show here). Of significance here is the fact that $\emptyset$ and $\mathbb{R}$ are (necessarily) open in both topologies. Now suppose that $V = (-\infty, a)$ for some $a\in\mathbb{R}$ is an arbitrary open set with respect to the lower topology. It is not too hard to see that $$f^{-1}(V) = f^{-1}((-\infty,a)) = (-a,\infty),$$ which is an open set in the upper topology. Thus if a set $V$ is open with respect to the lower topology, then the preimage of $V$ under $f$ is also open. Therefore $f$ is continuous.
• Continuity of $f^{-1}$: Finally, we need to show that $f^{-1}$ is continuous. But note that $f^{-1}(x) = -x$, so we can apply the same argument as above.

This finishes the proof, therefore $\mathbb{R}$ with the upper topology is homeomorphic to $\mathbb{R}$ with the lower topology, with the homeomorphism given by $x \mapsto -x$.