8

Definition of Topology

I don't understand this characterisation of a topology, I don't even comprehend what $g\colon\{ \{\{\}\},\{\{\},\{\{\}\}\}\} \rightarrow T$ should mean? Can somebody explain it to me?

Willie Wong
  • 73,139
StefanH
  • 18,086
  • 2
    If I read ${} = \varnothing = 0$ (as an ordinal), then it appears that the domain of $g$ is the Kuratowski ordered pair $(1,0)$, which equals the set ${1, 2}$. I don't know what it's supposed to mean, though. – Lord_Farin Feb 13 '13 at 13:07
  • 16
    A true masterpiece of obfuscation :-) – Martin Feb 13 '13 at 13:14
  • 2
    I think you should be glad you don't understand it. Otherwise it would mean you have a mind like the person that wrote that definition down! – Chris Godsil Feb 13 '13 at 13:37
  • 1
    This is just the kind of silliness that gives mathematics a bad rep! – Mozibur Ullah Jul 02 '13 at 11:28

3 Answers3

11

It is just an attempt to express the usual axioms of topology in a more "formal" manner. Like Lord_Farin wrote, we can write the sets $\mathbf{0} = \emptyset = \{\}$, $\mathbf{1} = \{\mathbf{0}\} = \{\{\}\}$, and $\mathbf{2} = \{\mathbf{0}, \mathbf{1}\} = \{\{\}, \{\{\}\}\}$ using von Neumann's notation for integers. Then the domain of $g$ is just the two-point set $\{ \mathbf{1}, \mathbf{2}\}$.

Now, $f:I \to T$ means that for each $i\in I$ we have $(i)f \in T$ is an element of the topology (here we follow the Wikipedia article to act functions on the right instead of on the left as usual). So $\cup_{I}(i)f \in T$ for any set $I$ and any function $f:I \to T$ is just a way to say, in notations, that "$T$ is closed under arbitrary union."

On the other hand interpreting $g: \{\mathbf{1},\mathbf{2}\} \to T$ and $(\mathbf{1})g \cap (\mathbf{2})g \in T$ just say, in notations, that "$T$ is closed under pairwise intersections".

Community
  • 1
Willie Wong
  • 73,139
  • Ah of course, that makes sense. However, we will probably never know why the postfix notation is used, and why not e.g. $\mathbf 0$ and $\mathbf 1$ are used in place of $\mathbf 1$ and $\mathbf 2$. Let alone the notation $\varnothing$... This way it's just needlessly confusing. – Lord_Farin Feb 13 '13 at 13:33
  • @Lord_Farin I think we can make an educated guess about the "why" from the Wikipedia history. See upcoming answer. – Erick Wong Feb 13 '13 at 13:45
  • In view of Erick's sleuthing, the OP (at Wikipedia) may have had in mind not von Neumann's notation, but that which was used in Morse-Kelley set theory. :-) – Willie Wong Feb 13 '13 at 14:27
  • Somebody removed the definition from Wikipedia. Maybe one should reproduce it here? – Martin Feb 13 '13 at 15:26
  • 2
    @Martin: the good thing about Wikipedia (which we reproduce here) is that all edits can be tracked in histories. So I've edited the OP to replace the link with one to a version where the definition stood. – Willie Wong Feb 13 '13 at 15:46
4

This is a bit hard to read when formatted as a comment, but here is the "motivation" in the words of the Wikipedia user who made that particular edit in October 2012.

Originally the definition was short and plain. Perhaps a bit quaint in the use of postfix, and subpar grammar and typography, but it isn't hard to piece together. The idea is to define a topology in terms of $T$ first rather than as a collection of subsets of $X$, which is not an unreasonable idea.

A topology is a set $T$ such that

  1. For all set $I$.For all function $f:I\to T$ $\cup(i)f \in T$.

  2. For all function $f:\{1,2\} \to T$ $(1)f\cap(2)f \in T$.

This was subsequently deleted by another user who described it as "an attempt to define $T$ as set of open sets in a most unreadable manner." The previous user then restored the content, adding the following message:

Hello,

First of all I want to say that the definition of a "topology" rather than a Topological space is standard (ie Kelley General Topology) The definition is correct and you can easily prove it The notation (a)f instead of f(a) is standard too and, imo, it emphasises the formal idea of a function, so to say (a)f is to say a in pi_1(f) where f:=(A,f,B) so, for example, b=(a)f is to say (a,b) in f. i corrected and supress the use of the non-necessary symbols 1,2

The current revision would appear to be merely a sarcastic response (temper tantrum?) to the criticism. The part that convinces me is how thoroughly he/she eliminated the use of "non-necessary" symbols 1,2, even from the enumerate list!

Erick Wong
  • 25,198
  • 3
  • 37
  • 91
  • Thanks for the comprehensive reply. It makes sense now. It is yet another example of the (un)reliability (mathematical) WP that this monstrosity has resided unchanged since Oct 2012. (IOW: If even this isn't changed, what other grave errors persist?) – Lord_Farin Feb 13 '13 at 14:02
  • 3
    @Lord_Farin: I wouldn't call it a "grave error". The mathematics is (not-so) obviously correct, and the point of view (defining a topology $T$ first then take the space as $\cup T$) is valid. An unfortunate choice of notations it may be, but certainly it is not factually wrong. – Willie Wong Feb 13 '13 at 14:24
  • 5
    I’m not sure that correct makes up for perversely unreadable! What on earth was wrong with saying that a topology is a set $\mathscr{T}$ such that for all $\mathscr{U}\subseteq\mathscr{T}$ and all $U,V\in\mathscr{T}$ we have $\bigcup\mathscr{U}\in\mathscr{T}$ and $U\cap V\in\mathscr{U}$?! – Brian M. Scott Feb 13 '13 at 23:50
  • @Willie I'm sorry, my choice of words was bad. My intended reading was: "If this isn't changed, then what other things, possibly grave errors, persist?" It's not what I wrote, but the point stands. – Lord_Farin Feb 16 '13 at 08:16
1

The set $\{0,1\}$ with the Sierpiński topology, which I will call $\rm S$ has the following property : there is a bijection between the topology $\mathcal T$ of any topological space $\rm X$ and the continuous functions $\rm X \to \rm S$.

Damien L
  • 6,649