Talk:Irreducible Space is Locally Connected
A basis is a subset of a topology, so all the sets comprising a basis are open.
If any two open sets have a non-null intersection, it follows that any two sets of a basis likewise have a non-null intersection, as those sets are also open. --18.104.22.168 04:08, 23 August 2011 (CDT)
That doesn't implie that there is a "local base" of connected sets. --Dan232 07:53, 23 August 2011 (CDT)
- That's where the problem is, the definition's wrong for Locally Connected. --22.214.171.124 08:02, 23 August 2011 (CDT)
- It should say, "local basis" instead of "basis". Either way, if any two sets of the basis have non-null intersection doesn't mean that they are connected (no trivially, at least) --Dan232 08:15, 23 August 2011 (CDT)
Nope, I feel a bit naive. I need to understand why:
- a) if any two sets are intersecting the space need not be connected, and
- b) you can always create a local basis for a point by taking the intersection of all the sets which contain $x$, and calling the set containing that set "a local basis". For sure, every open set containing $x$ will contain this set. Is that not a local basis? Is it not then this space always locally connected?
I'm missing something fundamental here which I can't find explained anywhere. --prime mover 16:01, 23 August 2011 (CDT)
- Regarding a), you don't have to check whether or not the space is conneted, but wether the sets of the basis as subspaces are connected or not.
- Regarding b), that is not a local basis. The infinite intersection of all neighbourhoods of $x$ may not be a neighbourhood of $x$, but it is contained in itself.
- Indeed every Hyperconnedted Space is Locally Connected, but I don't see this prove is enough. I think it needs more steps. By the way, I added an equivalence of locally connected for the use of local basis instead of basis in Equivalence of Definitions of Locally Connected Space --Dan232 16:18, 23 August 2011 (CDT)