Talk:Synthetic Basis formed from Synthetic Sub-Basis

From ProofWiki
Jump to navigation Jump to search

I had great difficulty getting the pages relating to Definition:Topology Generated by Synthetic Sub-Basis to fit together and flow properly. As such there is plenty room for improvement and I welcome any attempts to bring some shape to this. So my view is: have a go, see if you can make this section work. --prime mover (talk) 08:30, 23 September 2012 (UTC)

Why the change from A to F? Just curious, didn't seem to be consistent with general policy. Do we use F everywhere else in this area? --prime mover (talk) 16:31, 22 November 2012 (UTC)
I just didn't want to use the same symbol on Definition:Analytic Sub-Basis to define both $\BB$ and $\tau'$. (By the way, $\FF$ stands for "finite.") --abcxyz (talk) 16:36, 22 November 2012 (UTC)


I think there may be something wrong with this proof. A synthetic basis is defined to be a set of some sorts, but if the universe is an element of $\BB$, then $\BB$ can't be a set. Suppose that $\BB$ is a set, then by the axiom of unions, $\bigcup \BB$ is also a set. The universe isn't a set; a contradiction soon follows.

One way to fix this is to change the definition of a synthetic sub-basis so that it is also a cover for $X$, then change the construction of $\BB$ to

$\displaystyle \BB = \set {\bigcap \FF: \FF \subseteq \SS, \, \card \FF \in \N_{\gt 0} }$

Here we have the condition that the cardinality of $\FF$ is both finite and positive, so you won't be able to do anything weird like $\bigcap \O$.

Please let me know if I'm wrong or anything. --HumblePi (talk) 16:33, 13 January 2017 (EST)

First of all, sorry for the long delay... But to the content. The implicit assumption here is that we are working within $X$ as our universe where $\bigcap$ is concerned, so that $\bigcap \O = X$. I believe this addresses all your concerns.
We could argue whether this assumption needs to be explicated. I think it's pretty clear, but I'm also not opposed to doing so. However, changing the definition of subbasis would be overkill. — Lord_Farin (talk) 13:00, 24 April 2017 (EDT)
Thanks for the reply; I really don't mind if it takes a while.
I'm new to this idea of there being different universes, so I would just like to make sure I understand this. When you take a set $X$ to be the universe, that just means that your quantifiers ($\forall$ and $\exists$) range over $X$, and this is only done for the intersection $\bigcap \O$. Correct? --HumblePi (talk) 18:09, 24 April 2017 (EDT)
Exactly. And here it is taken implied because $\SS \subseteq \powerset X$. — Lord_Farin (talk) 11:35, 25 April 2017 (EDT)