Talk:Irreducible Space is Locally Connected

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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. -- 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. -- 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)