Talk:Partition topology

Check the usage of the separation axioms from COT are the same as those on wikipedia. jbolden1517^Talk 01:10, 12 June 2006 (UTC)

Sure about the odd-even topology? We have:

$P={\left\{\{2k|k\in \mathbb {N} \},\{2k+1|k\in \mathbb {N} \}\right\}}$

but I wonder whether it ought to be:

$P={\left\{\{2k,2k+1\}|k\in \mathbb {N} \right\}}$

As it stands, there is no indication that k needs to be the same for both the 2k and 2k+1 sets.

In fact, Steen and Seebach have it as:

$P={\left\{\{2k-1,2k\}|k\in \mathbb {Z} ,k>0\right\}}$

which emphasises and clarifies the specific range of k. It ultimately comes to the same thing, of course, but it allays the confusion over whether the conventional understanding of $\mathbb {N}$ includes zero or not. --Matt Westwood 22:55, 28 May 2011 (UTC)

Each point of X is a set in P?

The article says:

"The trivial partitions yield the discrete topology (each point of X is a set in P) or indiscrete topology (P = {X})."

What if the space is X=R^2 (it is not a space of sets, but of points), each point of X can't be a set in P, because it's elements are not sets.

This should be clarified. — Preceding unsigned comment added by 186.18.76.220 (talk) 20:46, 28 October 2011 (UTC)

Yes it can - each element of X is an ordered pair. So each set of the discrete topology contains a set of ordered pairs of X = R^2. — Preceding unsigned comment added by WestwoodMatt (talk • contribs) 23:03, 28 October 2011 (UTC)