# Why should a topological space itself be open?

For convenience, let $X$ be our space.

Specifically, can anyone name a few desirable properties or theorems that would fail if $X$ weren't required to be open? More generally, is there a part of topology that would completely fall apart?

It seems to me that we mainly want closure under arbitrary unions and finite intersections, which appears to be the more natural part of the definition (unlike "forcing" $\varnothing$ and $X$ to be open, which feels rather contrived). Of course, we get the empty set for free if we take an arbitrary union of nothing ($\bigcup \varnothing = \varnothing$), so that part really doesn't need to be in the definition.

Let's define a new word: A tolology on $X$ is a subset of $\mathcal P(X)$ that is closed under arbitrary unions and finite intersections. By the to(p/l)ological closure of a set I'm referring to the smallest to(p/l)ology containing it.

Let $\mathscr T$ be a topology on $X$ and consider $\mathscr T' = \mathscr T \setminus \{X\}$. If $\bigcup \mathscr T' = X$, then closure under arbitrary unions forces us to throw $X$ back in anyway, so the topological closure coincides with the tolological closure, nothing interesting here. Otherwise (this is what bothers me), we have $\bigcup \mathscr T' \subsetneq X$, then $\mathscr T'$ is still closed under arbitrary unions and finite intersections, so it's a tolology but not a topology. But throwing in $X$ adds nothing to the richness of the to(p/l)ology at all.

In fact, let's say $\bigcup \mathscr T' \subsetneq X$. Then the topological closure of $\mathscr T'$ is $\mathscr T$, but we still have a pretty boring space. For, if $\left|X \setminus \bigcup \mathscr T' \right| = 1$, then our space is not $T_1$, and if $\left|X \setminus \bigcup \mathscr T' \right| \geq 2$, then it's not even $T_0$. (Actually, any $T_1$ space must satisfy $\bigcup \mathscr T' = X$ by definition, and that probably covers just about every theorem in topology.)

Since un-requiring $X$ to be open doesn't give us any fewer theorems than we already have, and those spaces whose topology and tolology are different are as uninteresting as it gets, why can't we replace the definition of topology with that of tolology for simplicity's sake?

The only argument I can think of against this is that the first Kuratowski closure axiom says $\overline \varnothing = \varnothing$, so $\varnothing$ is closed, which means $X$ is open. But why do we need that first axiom?

• Just to point out, if the whole space is not open, then there exists a point which has no open neighborhood. Not sure if this is something undesirable. – William Jun 14 '12 at 7:14
• As you point out we get the empty set for free by considering an empty union. Similarly, we get the whole space by considering the empty intersection. It makes no sense to speak about tolology, it's exactly the same thing as topology. – in_mathematica_we_trust Jun 14 '12 at 7:21
• @chester: It's not controversial in this context, because we are working in the powerset of a fixed set. – Zhen Lin Jun 14 '12 at 7:41
• I'm with @ZhenLin. The empty intersection is well-defined because we're considering the powerset of a fixed set. – in_mathematica_we_trust Jun 14 '12 at 7:56
• Considering empty intersections, look at it from the point of view of axioms. Since topological spaces require only finite intersections of opens to be opens, the simplest and most natural axiom would be to require the intersection of two opens to be open. That gives you all finite intersections, except the empty intersection. You could add that one explicitly, which amounts to requiring $X$ open. Or you could instead replace "intersection of two opens" by "for a family of open sets, if the family is finite then the intersection is open" which is rather more involved to state. – Marc van Leeuwen Jun 14 '12 at 8:09

I think that it is reasonable to expect that a constant function from any space to any space would be continuous, and this of course true if and only if the all space is an open set.

Edit: I just realized that more is true: Suppose $X$ is a space in which $X$ is not open. Then there are no continuous maps from $X$ to spaces in which the all space is open, because if there is such a map $f:X\to Y$, then we must have $f^{-1}(Y) = X$ is open, which is false.

• But why should there be a continuous map from $X$ to $Y$ where $Y$ is open? – user33661 Jun 14 '12 at 7:51
• Your first statement is not true: the constant function form any space to another, whose constant value is a point not contained in any open set (which exist in tolologies that are not topologies) would be continuous. – Marc van Leeuwen Jun 14 '12 at 7:56
• Marc van Leeuwen, what I meant is: all constant maps from $X$ are continuous if and only if $X$ is open in $X$. – the L Jun 14 '12 at 7:58
• chester, category theory teaches us that one should study an object by morphisms to and from this object. On the one hand, your objects do not a rise naturally in "nature". On the other hand, there is no interaction between your objects and the classical objects. So why should we study you objects? – the L Jun 14 '12 at 8:01
• @anonymous: which as I said is false. Let $p_0\in X$ be a point not contained in any open set, then (1) $X$ is not open in $X$, and (2) the constant function $X\to X$ defined by $x\mapsto p_0$ is continuous. – Marc van Leeuwen Jun 14 '12 at 8:02

You get $X$ for free from closure under finite intersections: $\varnothing$ is a finite collection of open sets $-$ it’s certainly finite, and it doesn’t contain any non-open sets!) $-$ and

\begin{align*} \{x\in X:x\in\bigcap\varnothing\}&=\{x\in X:\forall S\in\varnothing(x\in S)\}\\ &=\{x\in X:\forall S(S\in\varnothing\to x\in S)\}\\ &=X\;. \end{align*}

But quite apart from such technicalities, carrying along the dead weight of points in $X\setminus\bigcup\mathscr{T}'$ would unnecessarily complicate the statements and proofs of a large enough number of results to be a real nuisance. And there are areas of mathematics in which spaces that aren’t $T_1$ are of real interest.

• See this regarding "a large number of results" and "spaces which aren't $T_1$". – user33661 Jun 14 '12 at 7:55
• @chester: Your first comment is almost certainly wrong: any argument that contains something like ‘Let $x\in X$ be arbitrary’ is very likely to need modification: there’s a fundamental distinction between points that have nbhds and points that don’t. It’s perfectly true that non-$T_0$ spaces are rather seldom dealt with, but I have run into applications of them. – Brian M. Scott Jun 14 '12 at 8:31

A tolological space $(X,\mathscr T')$ would be exactly a topological space $(X',\mathscr T')$ with $X'=\bigcup \mathscr T'$ plus a set $S=X\setminus X'$ of points not contained in any open set. (Which points need to be distinguished from the possible points of $X'$ not contained in any open set except $X'$.) The tolology says nothing about the points of $S$, and exactly because this is pretty boring one gains very little by allowing the possibility of such points; they will just possibly get in the way for some arguments. And if you argue that most interesting statements require separation axioms rather stronger than requiring every point to be contained in some open set, then this is not really an argument against including the weaker version among the basic axioms of topology.

If you take any subset $X$ of a topolgical space $T$, and $\mathscr T'$ is the part of $\mathcal P(X)$ consisting of sets that are open as subsets of $T$, then $(X,\mathscr T')$ is a tolological space, so you still can study these beasts in the context of topological spaces if you like.

Throwing out the axiom that $X$ is open might simplify the axioms a bit, but it might complicate the statement and proofs of a great many theorems. Admittedly, it will be a minor complication, but multiplied over a huge number of results it would be a major pain in the posterior. And all that for no actual gain, as the extra generality is really quite trivial.

• But it only complicates the statements and proofs of the theorems involving non-$T_0$ spaces. And the only $T_0$-but-not-$T_1$-spaces whose theorems need to be modified are the spaces with exactly one point with no neighborhood (other than $X$). I'm not sure there are too many theorems in these spaces. Besides, I'm of the (possibly wrong) impression that virtually no topologist considers non-$T_0$ spaces. – user33661 Jun 14 '12 at 7:36
• I think it is more accurate to say it only complicates statements and proofs of theorems that might possibly involve non-$T_0$ spaces. And that is a much bigger class than those that explicitly mention such spaces. – Harald Hanche-Olsen Jun 14 '12 at 9:35

I have seen consideration of Kuratowski "closure" axioms. Omit the one that says the closure of the empty-set is empty, and you get an alternate to topological space. (So that the whole space may not be open.) As I recall, the corresponding change when you consider convergence of nets is merely that you allow an empty directed set. The set of limits of the empty net is then that closure of the empty set.

For open sets, we have: if $A,B$ are open then $A \cap B$ is open. But it does not follow that any finite intersection of open sets is open. Merely that any finite intersection of one or more open sets is open... The intersection of zero open sets need not be open.

There is a point of view where what is important about topology is precisely the open sets -- points are only an artifice to make it easier to talk about spaces, open sets, and functions easier to talk about. From this point of view, having extra points which are completely and utterly separate from the open sets is rather unnatural.

One can even eschew points entirely and talk about frames and locales. This is whimsically called "pointless topology". There is some reason to believe that this might gives a better notion of topology than the usual one.

Some pragmatic issues: There would be questions about topologies defined through their closed sets. I would find it much more difficult to prove that the spectrum of an integral domain is irreducible. There is also a beauty to the spectrum having both the open and the closed semantics. You now now create subspaces with less well-structured set systems, and you'd have to check whether you can create a subspace topology that preserves the open/closed subspace distinction. If you can, does it provide descriptive value? Is it worth losing the generation of structure imposed on taking a subset as a subspace? Discrete subgroups will get iffy.