I see a little more in the question now. It seems that the OP is proposing to eliminate the axioms that the empty space and the total space are open but maintain the axioms that arbitrary (nonempty!) unions and finite (nonempty!) intersections of open sets are open.

In this case there is a little content here, because you can try to figure out whether or not $\emptyset$ and $X$ will then be open or not.

Note one disturbing fact: with the elimination of the above axiom, there is nothing to imply the existence of any open sets at all! Whether this is good or bad, if it happens there is nothing further to say, so let's assume that there is at least one open set.

Claim: If $X$ is a Hausdorff with more than one point, then the empty set and the total space are open.

Proof: Indeed, the Hausdorff axiom asserts that any two points have disjoint open neighborhooods, so the intersection of these is empty. The same axiom says in particular that every point has at least one open neighborhood (!) which is clearly equivalent to
$X$ being open.

Claim: If $X$ is T1 with more than one point, then the total space is open, but the empty set need not be.

Proof: T1 means that the singleton sets are closed. If there are least two of them, take
their intersection: this makes the empty set closed, hence the total space open. On the other hand, the cofinite topology on an infinite set is T1 and the empty set is not open (if we do not force it to be).

I am coming around to agree with K. Conrad in that having $X$ be open in itself may not be just a formality. A topological space in which some point has no neighborhood sounds like trouble...I guess I was thinking that if you get into real trouble, you can just throw $X$ back in as an open set! If you want to put it that way, this is some kind of completion functor from the OP's generalized topological spaces to honest topological spaces.

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