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A topological space is a set $X$ together with a topology $\tau$ (a collection of open subsets) such that.

  1. $\emptyset\in \tau$ and $X\in \tau$.
  2. The intersection of a finite number of sets in $\tau$ is also in $\tau$.
  3. The union of an arbitrary number of sets in $\tau$ is also in $\tau$. (it could be the union of infinitely many sets)

what if we tweak this definition so that it becomes:

A topological* space is a set a set $X$ together with a topology* $\tau$ (a collection of open subsets) such that.

  1. $\emptyset\in \tau$ and $X\in \tau$.
  2. The intersection of an arbitrary number of sets in $\tau$ is also in $\tau$.
  3. The union of a finite number of sets in $\tau$ is also in $\tau$.

how will that change how the concept of a topology captures that of a neighborhood intuitively speaking? (this is the most important to me) how are these topological* spaces going to differ from topological spaces? was this particular definition ever historically coined down?

thx in advance for any kind of help

user153330
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2 Answers2

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If $\{C\}$ is a collection of sets satisfying the requirements of your tweaked definition, then $\{X-C\}$ is an ordinary topology in the original sense. In other words, your "tweaked" definition produces the closed sets of some topology consisting of open sets in the ordinary sense.

Lee Mosher
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  • could you give more details please? do you mean by the notation ${X-C}$ that ${X-C_1,X-C_2,...,X-C_n}$ where the $C_i$ are the sets in the collection ${C}$? – user153330 Feb 06 '16 at 15:08
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    That is exactly what I mean, except for the fact that these sets need not be finite. So one could use indexing to write ${C_i}{i \in I}$ and ${X-C_i}{i \in I}$. – Lee Mosher Feb 06 '16 at 15:15
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    User, why don't you try to prove it yourself? Let $\tau$ be a collection of subsets of a set $X$. Then $\tau$ is a topology* if and only if ${ U : X - U \in \tau}$ is a topology. – GEdgar Feb 06 '16 at 15:15
  • @GEdgar okay, i will try to work out a proof myself, but i'm also interested into learning why this definition wont work *intuitively* speaking – user153330 Feb 06 '16 at 15:17
  • This definition will work. It defines the collection of closed sets, just as the conventional topology defines the collection of open sets. – GEdgar Feb 06 '16 at 15:18
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    The intuition that you seek is simply an intuitive understanding of the difference between open and closed sets in an ordinary topology. – Lee Mosher Feb 06 '16 at 15:18
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    @user153330: Just note that $$X\setminus\varnothing =X$$ $$X\setminus X=\varnothing$$ $$\bigcap_{C\in\mathscr C} (X\setminus C)=X\setminus\bigcup_{C\in\mathscr C}C$$ $$\bigcup_{n=1}^N(X\setminus C_n)=X\setminus\bigcap_{n=1}^N C_n$$ – MPW Feb 06 '16 at 15:19
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It is clear that items 1 and 3 in your enumeration are included in (implied by) the original definition of topological space; hence these are natural properties that a “neighborhood” should have. So the only point to discuss is whether lifting the restriction of finite intersections takes somewhere counterintuitive.

Remark. Neighborhoods of a point $x$ are usually sets $V$ such that there exists an open $U$ with $x\in U \subseteq V$, so one should take into account that the intuition is about subsets that are neighborhoods of all of their points.

A first example: The discrete topology $\mathcal{P}(X)$ on any $X$ satisfies both the axioms for a topology and yours. This is a first hint that anything one can say should be based in some strong picture of neighborhood, like that of a (locally) connected space as $\mathbb{R}^n$. Even “metric space” is not enough, since the discrete topology is metrizable.

Take, for instance, the concept of a limit of a sequence. We would like to say that $x_n\to x$ as $n\to\infty$ whenever the tail of the sequence $\{x_n\}_{n>N}$ is “closer” to $x$ as $N$ grows. That would mean, taking neighborhoods $U_1\supset U_2\supset U_3\supset \dots \supset U_m\supset \dots$ of $x$, all but finitely many terms of the sequence lie in $U_1$, and the same is true for the other $U_m$. If you allow arbitrary intersections, there must be a smallest neighborhood of $x$, and hence in a topological* space the sequence gets infinitely close to $x$ in a finite time. That is, there is an $M$ such that for all neighborhoods $U$ of $x$ and all $n>M$, $x_n\in U$.

If in your intuition of space, you may approach a point without never being “infinitely close” but getting every time closer, you can't have arbitrary intersections of neigborhoods.

EDIT: As Paul Sinclair observed in the comments, allowing topologies* that are not closed under arbitrary union also goes against some intuitions. This follows easily from the highlighted remark above, plus the observation that (using that terminology) if $V$ is a neighborhood of $x$ and $W\supseteq V$, then $W$ also is. This automatically implies closure under arbitrary unions.

Pedro Sánchez Terraf
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  • You missed that he/she also changed (3) to requiring only finite unions. That is why as Lee Mosher said, it now describes the coTopolgy of closed sets instead of the Topology of open sets. – Paul Sinclair Feb 06 '16 at 22:19
  • @PaulSinclair No, I didn't miss that. I wanted to make a point around arbitrary intersections, allowing (at least) finite unions doesn't go against any "neighborhood" intuition. I actually upvoted his answer. – Pedro Sánchez Terraf Feb 06 '16 at 22:40
  • Then you are mistaken in saying in the first paragraph that the only point to discuss is lifting the restriction of finite intersections. The allowing of *topologies that do not allow infinite unions also has an impact. – Paul Sinclair Feb 06 '16 at 22:59
  • @PaulSinclair Ok, that's fine. It is directly related to my highlighted remark. I'll edit the answer as soon as I get to my laptop. – Pedro Sánchez Terraf Feb 06 '16 at 23:12
  • I figured as much, but that was why I had assumed you had missed the change in (3) from the normal axiom. – Paul Sinclair Feb 07 '16 at 00:18
  • @PaulSinclair Hope my editing satisfies both you and the OP! – Pedro Sánchez Terraf Feb 07 '16 at 12:22