I'm wondering if the following can be true: Let Y be a second countable space and $\pi_2:Y \times \mathbb{R}\rightarrow\mathbb{R}$ ($\mathbb{R}$ with its usual topology and $\pi_2$ the projection onto the second factor) be a closed map: do these assumptions imply that Y is compact? (There is no assumption $T_0$, $T_1$ or $T_2$ on $Y$.)

thank you in advance.


Let $(y_n)$ be a sequence in $Y$. Let $A$ be the subset of $Y \times \mathbf{R}$ of all points $(y_n, \frac{1}{n})$ for $n \in \mathbf{N}$, and let $B$ be its closure. Then $\pi_2[B]$ is closed in $\mathbf{R}$, and contains all points $\frac{1}{n}$, so it contains $0$. So for some $y \in Y$, $(y,0) \in B$. Using the countable base we can extract a subsequence of the $(y_n)$ that converges to $Y$ (as $Y$ is first countable in particular). We do then need that $y$ is in the closure of all subsequences of $(y_n)$ as well, which follows in a similar way, otherwise we cannot get (without separation axioms) a convergent subsequence from first countability alone. But this works.

So $Y$ is sequentially compact, which implies that $Y$ is countably compact (in the covering sense; no separation axioms needed) and as $Y$ is also Lindelöf, being second countable, $Y$ is compact.


I assume that by compact, you mean quasi-compact (i.e. not necessarily Hausdorff but with the finite sucover property), otherwise any $Y$ with the coarsest topology would be a counterexample.

Then it seems that your notion is equivalent to pseudocompactness, namely that any continous $f:Y\to\mathbb{R}$ has compact image. I checked that quickly, so I may be wrong.

EDIT : indeed I am, see below. The property is slightly stronger than that.

(I haven't found a reference in MR, although I would bet it is an exercise in Bourbaki).

The idea is first to observe that the property remains the same if you substitute $\mathbb{R}$ with $[0,1]\simeq[-\infty,\infty]$, basically because any one is included in the other (up to homeo).

Then since the graph of a continuous $f:Y\to\mathbb{R}$ is closed, $f(Y)$ must be closed in $\mathbb{R}$ and also $[-\infty,\infty]$, hence bounded. Similarly, any lower/upper semi-continuous $f$ is lower/upper bounded and attains its inf/sup, since $f$ is lsc iff its "epigraph" $\{(y,t):f(y)\leq t\}$ is closed.

For the converse, if $F$ is closed in $Y\times[0,1]$, but not its projection $F_2$, one may assume that $0$ isn't in $F_2$ but is in its closure. Then the function $y\mapsto \inf t : (y,t)\in F $ is lower semi-continuous and doesn't attain its infimum $0$.

EDIT: pseudocompactness doen't imply that lsc functions attain their infimum in general (not completely regular) spaces : there is no way to construct a continuous function from an lsc one if there are "not enough" continuous functions. Your property is indeed stronger than pseudocompactness, as the particular point topology on $\mathbb N$ shows. This space is homeomorphic to $\mathrm{Spec}(\mathbb{Z})$ with Zariski topology : a discrete countable subspace (primes) plus a dense "generic point" (0). Any continuous function is constant (hence pseudocompactness), but a function $f$ is lsc iff $f\leq f(0)$ !! (I checked that twice, it's so shocking!)

So your property is equivalent to "lsc real-valued functions attain their lower bound", which may have a name, I don't know. Maybe "strongly pseudocompact" ?

By the way, pseudocompact doesn't imply (quasi-)compact, even for "nice" (completely regular) spaces, as is seen with the long line or simply the first uncountable ordinal $\omega_1$ (with order topology). There are also non-Hausdorff examples, see wikipedia article.

  • $\begingroup$ I thought about the latter too, as all countably compact space satisfy the fact that the projection onto R is closed (we can replace R by any sequential space even). But of course I realized we cannot have a second countable (note that condition! the OP works in second countable spaces) counterexample, as a second countable space is Lindelöf, and a Lindelöf and countably compact is compact... So the standard examples do not work. I also don't see why we can replace R by [0,1]: closedness is not preserved by embedding... So please elaborate on that? $\endgroup$ – Henno Brandsma Jul 13 '10 at 16:38
  • $\begingroup$ This is the following lemma : If projection $Y\times Z_1\to Z_1$ is closed and $Z_2$ is a subspace of $Z_1$, then $Y\times Z_2\to Z_2$ is also closed. Proof: take $F$ closed in $Y\times Z_2$ the projection $G$ of its closure $\overline{F}$ in $Y\times Z_1$ is closed in $Z_1$, so $G\cap Z_2$ is closed in $Z_2$. But since $\overline{F}\cap(Y\times Z_2)=F$ (opens of subspace are traces of opens of ambient space), this is the projection of $F$. $\endgroup$ – BS. Jul 13 '10 at 18:31
  • $\begingroup$ And I overlooked the "second countable" hypothesis, which ruins my counterexamples. $\endgroup$ – BS. Jul 15 '10 at 9:50

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