show/hide this revision's text 2 added 1047 characters in body; added 3 characters in body

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.

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 lsciff its "epigraph" ${(y,t):f(y)\leq t}$ is closed.

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" ?

show/hide this revision's text 1

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 (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.

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 $0$.

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.