Are Closed Subspaces of Pseudocompact spaces Pseudocompact? Are Closed Subspaces of a Pseudocompact Space also Pseudocompact? I suspect this is the case, but I haven't been able to prove this.
 A: Take $\mathbb{R}$ with the topology whose open sets are sets with countable complement, name it $(\mathbb{R},\tau)$, and let $\mathbb{R}$ have the standard topology. Note that every two non empty open sets of $(\mathbb{R},\tau)$ have non empty intersection, in particular it is connected. Let $f : (\mathbb{R},\tau) \rightarrow \mathbb{R}$ be continouos. Soppose that $a \neq b \in imf$, then $(a,b) \subset imf$, take $c \in (a,b)$, then $f^{⁻1}(a,c) \cap f^{-1}(c,b)= \emptyset$, thus $imf$ is a singleton, and therefore  $f$ is bounded. Now take $\mathbb{N} \subset (\mathbb{R},\tau)$ it is closed, and it inherits the discrete topology, indeed $\{n\}= (\mathbb{R}\setminus\mathbb{N} \cup )$ so $i: \mathbb{N} \hookrightarrow \mathbb{R}$ is continuous.
A: No, the Mrówka $\Psi$-space (see e.g. here) is pseudocompact, locally compact, completely regular (Including Hausdorff), but has an uncountable closed and discrete subspace (all points outside $\omega$, the open and dense discrete space) which is very much non-pseudocompact.
A normal (or $T_4$) pseudocompact space is countably compact and so its closed subspaces are also countably compact (and hence pseudocompact too). So completely regular ($T_{3\frac12}$) is the best we can do, separation-axiomwise.
A: If the space is compact, then closed subspaces are pseudocompact since their closed subsets are compact and so map to bounded images.
Otherwise, the answer is not necessarily. Consider the set $\mathbb{N}$ with the (particular point) topology of open sets $U$ containing $0$. It is pseudo-compact because the only continuous functions are constant. But the closed subset $\mathbb{N}\setminus\{0\}$ is not pseudo-compact because it inherits the discrete topology and so all functions on it are continuous, even the embedding map.
