Talk:Separated sets

Not Expandable!

Hi, it seems that this article cannot be more than some kind of a disambiguation page for the "separation" concept in topology. Anybody has any idea? 🤔 Mojtabakd «talk» 08:16, 29 May 2021 (UTC)[reply]

Why change the staggered formatting of the various definitions?

@Mgkrupa: Why did you change the staggered formatting of the various properties in the Definitions section? It was really nice to visually indicate that each property was a strengthening of the previous ones. PatrickR2 (talk) 00:14, 12 January 2023 (UTC)[reply]

Because it is not not encyclopedic and goes against Wikipedia's manual of style. See MOS:LISTBULLET and Wikipedia:Manual of Style/Lists. Mgkrupa 00:25, 12 January 2023 (UTC)[reply]

If it was a regular bullet list, I would agree with you. But the previous layout was more than a simple list. It was also intuitively conveying in a visual manner the relationship between the properties. PatrickR2 (talk) 00:59, 12 January 2023 (UTC)[reply]

Incorrect definition of sets separated by a continuous function

It seems that the definition of separation by a continuous function has a problem.

It says A and B a separated by a continuous function iff $f[A]=\{0\}\wedge f[B]=\{1\}$ . This means that $\emptyset$ can never be separated by a continuous function because for any function f $f[\emptyset ]=\emptyset$ .

This in turn means that for any topological space X, the whole space X and $\emptyset$ are never separated by a continuous function.

At the same time X and $\emptyset$ are precisely separated by a continuous function. To show that just take a constant function f, such that $(\forall x\in X)(f(x)=1)$ . Obviously $f^{-1}[{0}]=\emptyset$ and $f^{-1}[{1}]=X$ , so $\emptyset$ and X are precisely separated by a continuous function.

So we have two sets which are precisely separated by a continuous function but not separated by a continous function. This contradicts the article on separated sets which says "The properties are presented in increasing order of specificity, each being a stronger notion than the preceding one."

The easiest way to fix this contradiction is to amend the definition of "separated by a continuous function" to:

A and B a separated by a continuous function iff there exists $f:X\to \mathbb {R}$ such that $A\subseteq f^{-1}[{0}]$ and $B\subseteq f^{-1}[{1}]$ .

(I just checked the ncatlab site and this is the definition that they use for separation by a continuous function.)

Or alternative but equivalent condition would be: $f[A]\subseteq \{0\}$ and $f[A]\subseteq \{1\}$ .

Unless there are objections or other suggestions I will change the definition of "separated by a continuous function" to the first proposed one. Vesselin.atanasov (talk) 07:19, 31 January 2023 (UTC)[reply]

You are right that there is a slight problem here. Your suggestion is fine. To make it clear for novices of topology, you could even say something like:

A\subseteq f^{-1}(0)

and

B\subseteq f^{-1}(1),

that is, members of

A

map to

0

and etc...

(note: writing

f^{-1}[\{0\}]

is unnecessary formality. I know square brackets can be useful in set theory to make a distinction between images or inverse images of sets versus elements, but let's keep things simple). PatrickR2 (talk) 16:29, 31 January 2023 (UTC)[reply]

And then should we also flip the order of the equality in the next paragraph to

A=f^{-1}(0)

for symmetry? PatrickR2 (talk) 16:31, 31 January 2023 (UTC)[reply]

I agree that flipping the order of the equality makes sense because it will be easier for the reader to compare the definitions.

I will make these changes now. Vesselin.atanasov (talk) 11:05, 1 February 2023 (UTC)[reply]

A single closed neighborhood

If I am not too doddering in my thinking, an equivalent condition to separated by (open) neighborhoods, is the use of a single closed neighborhood. That is, sets $A$ and $B$ are separated by neighborhoods if and only if $A$ has a closed neighborhood $C$ for which $C \cap B = \emptyset$ . (Stating the obvious: that $C$ is a neighborhood of $A$ means there is an open set $U$ for which $A \subseteq U \subseteq C$ . That $C$ is closed and does not intersect $B$ means that $V$ , equal to the complement of $C$ , is open and $B \subseteq V$ and $U \cap V = \emptyset$ . Going the other way on the if and only if, just set $C$ to be the complement of $V$ .)

If there are textbooks that use this formulation of "separated by neighborhoods", let's add it to the article. — $Q$ uantling (talk | contribs) 17:48, 3 July 2023 (UTC)[reply]

The fact that you mention is true. It does not seem necessary to add it to the article though as it would break the focus of the article, which is to compare the various notions in which two sets can be "separated" and to show there is a natural progression in the definitions. Adding various equivalent conditions for each condition is not what this article is about, neither would be adding statements of a bunch of results involving these notions. Really the most natural definition for "separated by neighborhoods" is that the two sets have disjoint neighborhoods, and we should keep it at that. PatrickR2 (talk) 21:59, 3 July 2023 (UTC)[reply]

The approach I mention makes the transition from T_1 to T_2 more seamless IMHO; i.e., where T_1 has an open neighborhood, T_2 has a closed neighborhood. I like the flow as you describe, but I also want to think about this some more and see if I can come up with something that hits both priorities. —

Q

uantling (talk | contribs) 23:15, 4 July 2023 (UTC)[reply]