Do the empty set AND the entire set really need to be open? [closed]

Question

My question is motivated by the previous discussion 'Why is a topology made of open sets?'. While the axioms for arbitrary unions and finite intersections are without doubt essential to the concept of a topological space, the 1st axiom (the set itself and the empty set are open) seems rather technical. So, do we really need these conditions in order to build most (if not all) of the point-set-topology without significant changes? In other words, if we leave out the 1st axiom, can point-set-topology stil remain as useful and powerful in terms of what we actually need to do analysis and geometry?

EDIT: Expanded the topic name in accordance with my question.

Answer 1

Here's a boring reason, and it may or may not convince you: any function $f : X \to Y$ between topological spaces has the property that the preimage of the entire space $Y$ is the entire space $X$, and the preimage of the empty subset of $Y$ is the empty subset of $X$. So if you allow topological spaces in which either the entire set or the empty set is not open, there are no continuous functions from these spaces to "classical" topological spaces! Given that you agree with me that this is undesirable behavior, I think you are forced to make the entire set and/or the empty set either always open or always not open, and I think if you pick the second option then nothing changes except that, as KConrad says, it becomes unnecessarily harder to say things.

Actually, the situation is even worse: if the empty set isn't allowed to be open in $X$ then the continuous functions $X \to Y$ cannot miss any open set in $Y$, and if the entire set isn't allowed to be open in $X$ then the continuous functions $X \to Y$ cannot take values entirely in a proper open subset of $Y$. I think these are both much more unnatural than allowing the entire set and the empty set to be open. This is assuming you agree that the standard definition of continuity is natural.

Answer 2

If the empty set and the whole space are not open, then many statements you would like to make about open sets need qualifying remarks. It really can happen that two open sets are disjoint (two open balls that are far apart) or their union is the whole space (an appropriate pair of open half-planes that overlap). If the empty set were not open then we would have to say that any finite intersection of open sets is open or is empty. You'd have to tack on "or empty" in a lot of statements (e.g., the complement of a closed set is open or is empty... I assume you would like to call the whole space closed?). It is easier to allow the empty set as an open set to avoid a profusion of "or empty" qualifiers in theorems.

If, as has been suggested in a comment, the issue being raised is whether or not that first axiom about topologies is simply redundant, it isn't. Without that axiom we could consider any single subset of a space as a topology on the space: that one set is closed under arbitrary unions and finite intersections of itself. In that setting the concept of an open cover loses its meaning, so it really seems like a dead end.

Edit: Without the whole space being allowed as open (which can happen for "topologies" without that first axiom), there need not be open coverings, and then the usefulness of point-set topology is seriously damaged.

Answer 3

If one wants constant functions to always be continuous, then one must necessarily have the empty set and the whole space to be open.

From a category theory perspective, it is the continuous functions that are the more fundamental building block of topology, than the open or closed sets. I believe that there is some equivalent way to axiomatise topology via continuous functions using the machinery of sheaves, which is in some ways more "natural" than the simple but somewhat arbitrary-looking axioms for open sets, but I am not an expert on these matters.

Answer 4

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.

Answer 5

This question may be better answered by going to the intuition.

Let X be a set with some sort of "local structure"; a metric space will do, but any reasonable sort of "closeness" is fine. We want to say a set U ⊆ X is open if, whenever U contains a point p, it also contains all those points q ∈ X which are "near p".

Given this, U = X is clearly open. If p ∈ U, we want all the points of X "near p" to be in U; but every point of X is in U, so this is trivially true.

∅ is also clearly open. For this to be false, ∅ would have to contain a point p but not all the points near p; but as ∅ contains no points, this is impossible.

Also consider the following: Let f : X → {0,1} be the constant function sending every p ∈ X to 0. We want constant functions to be continuous, no matter how fine the topology on {0,1}. So f ^-1[{0}] = X and f ^-1[{1}] = ∅ both must be open (and both must be closed).

Answer 6

To address the question in a somewhat less-categorical way, I would point out that you can in fact do all of topology without using the expression "open set", by instead refering to the filter of neighborhoods of every point --- then a function $f:X\rightarrow Y$ is continuous iff for all $x\in X$ and every neighborhood $V$ of $f(x)$ there is a neighborhood $U$ of $x$ such that $f(U)\subset V$. You'll remember this as the "epsilon-delta"-style of continuity criterion from calculus, only without mentioning $ε$ or $δ$. The viewpoints are equivalent in the sense that one can define an open set as being a neighborhood of all its points, or contrariwise define a neighborhood of $x$ as including some open set containing $x$.

There's another alternative, to study presentations of topologies; usually a base for a topology is given. These have the advantage of being available as raw data, because the closure requirements for a base are much less stringent from a set-theoretic point of view than those of the whole system of open sets; one consequence is that you can study the topological space $(X,\langle B\rangle)$ in any universe including $X$, $B$ and $(X,B)$. This notion of a presented topology then lets you compare how the properties of a space-given-the-base change after a forcing extension of the universe --- if you're into that sort of thing.

The moral of this story is that open sets are not really the object of study in topology, and changing what you mean by "open set" is mostly going to distract you; continuous functions are one object of study --- as others have pointed out --- and there are many ways to define those.

Answer 7

One can define a topological space in terms of closed sets to make this easy to see.

A topological space $T$ satisfies:
- The intersection of any sets $X_i \in T$ is closed (i.e., in $T$)
- The union of finitely many sets $X_i \in T$ is closed.

Then, if we take, e.g., $T=\mathbb{R}$, by the first property above, $(0,1) \cap (10,20) = \emptyset \in T$. So we should consider the empty set to be closed.

The complement of $\emptyset$ is $\mathbb{R}$, so the whole space is closed as well (there was clearly nothing particular about using the reals here so this holds for arbitrary $T$). So we should consider $T$ to be closed as well. [No, you have used the existence of disjoint open sets, which does not hold in a general topological space -- PLC.]

By De Morgan's laws, one can check that the above characterizations of a topological space are equivalent to the usual definition in terms of open sets, so for consistency we should require $\emptyset$ and $T$ to be both open and closed in addition to the above two properties.