$\mathbb R$ with the right topology generated by $\tau = \{(a, \infty)\}$ is pseudocompact: proof by contradiction in terms of *open sets* I'm trying to prove that the topological space $X$ that is basically $\mathbb R$ equipped with the right topology generated by $\tau = \{(a, \infty): a \in \mathbb R\}$ is pseudocompact (any continuous function $f: X \to \mathbb R$). This question has been asked previously and also been answered but here I'm specifically looking for a review of my specific approach towards a proof.
This answer by Severin Schraven does a proof by contradiction in terms of closed sets. I want to do the same proof in terms of open sets, i.e., using the property that preimages of open sets under continuous functions are open.
My Approach:
Note that an open set in $X$ is of the following forms:
$$\emptyset, \quad (-\infty, +\infty), \quad (a, \infty).$$
Now suppose we pick some $x \in \mathbb R$ and look at the union of the disjoint open sets in its complement $\mathbb R \setminus \{x\}$, that is, $(-\infty, x)\cup (x, \infty)$. In the standard topology on $\mathbb R$, the sets $(-\infty, x)$ and $(x, \infty)$ are certainly both open and disjoint.
We also know that it is an usual property of mappings that $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$.
So $$f^{-1}(-\infty, x) \cap f^{-1}(x, \infty) = f^{-1}((-\infty, x) \cap (x, \infty)) = \emptyset.$$
This implies either $f^{-1}(x, \infty) = \emptyset$ or $f^{-1}(-\infty, x) = \emptyset$ or both of them are $\emptyset$. In fact, to prove that $f(X) = x$, that is $f$ is a constant map, we need to prove that both of the pre-images are empty, i.e., $f^{-1}(x, \infty) = \emptyset$ as well as $f^{-1}(-\infty, x) = \emptyset$.
After this, I was thinking of picking a $y \in \mathbb R$ such and looking at $f^{-1}(\mathbb R \setminus \{y\})$ to show that it is in fact not possible for $f^{-1}(\mathbb R\setminus \{x\})$ to be non-empty by producing some contradiction. That is neither $f^{-1}(-\infty, x)$ nor $f^{-1}(x, \infty)$ are allowed to be empty due to some resulting contradiction. But I'm not sure how to go about it. Can this be shown by contradiction, similarly to Severin's approach?
Certainly, any proof regarding continuous functions can be done in terms of open sets as well as in terms of closed sets, and such proofs are supposedly "dual" in some sense. I'm basically looking for a version of Severin's proof in terms of open sets.
 A: The right topology has the properties that

*

*all non-empty open sets intersect (anti-Hausdorff, or hyperconnected.

*all non-empty closed sets intersect (or ultraconnected).

For both of these kinds of spaces $X$ we have that all continuous $f: X \to \Bbb R$ are constant.
The usual arguments given in the linked answers focus on 1, and note that if $f$ is not constant, there are two distinct values, which have disjoint open neighbourhoods $U,V$ in $\Bbb R$. Then $f^{-1}[U]$ and $f^{-1}[V]$ are also disjoint (set theory, as $f^{-1}$ preserves intersection, as you note) and non-empty (as $U$ and $V$ contain values of $f$).
So those arguments can be generalised to

If $f: X \to Y$ is a continuous map from a hyperconnected space $X$ to a Hausdorff space $Y$, $f$ is constant.

Severin's argument is slightly different: it uses that all $\{x\}$ are closed in $\Bbb R$ instead. All sets $f^{-1}[\{x\}]$ for distinct $x$ are disjoint, and non-empty iff $x$ occurs as a value. So his argument can be summarised as

If $f:X \to Y$ is a continuous map from an ultraconnected $X$ to a $T_1$ space $Y$, $f$ is constant.

I wouldn't necessarily call these proofs dual. For that we'd have to use sets $\Bbb R\setminus \{x\}$ instead and use finite unions instead of finite intersections. From a general view they go for slightly different results, with similar proofs. The actual dual would be something like this:
Suppose $f: X \to \Bbb R$ is continuous and not constant, and has values $y_1= f(x_1) \neq f(x_2)= y_2$. Then $f^{-1}[\Bbb R \setminus \{y_1\}]$ is open (continuity), is non-empty (as $x_2$ is in it) and not $X$ (as $x_1$ is not) and similarly for $f^{-1}[\Bbb R \setminus \{y_2\}]$.
But $$X =  f^{-1}[\Bbb R \setminus \{y_1\}] \cup f^{-1}[\Bbb R \setminus \{y_2\}]$$
and so we've written $\Bbb R$ in the upper topology as a union of two open sets, none of which are $\Bbb R$. This cannot happen as $(a,\infty) \cup (b, \infty) = (\min(a,b), \infty) \neq \Bbb R$ for any $a,b$.
Well, I think the horse is now well and truly dead..
