|
|
Re: definition of closure in topological space question
Posted:
Nov 17, 2012 9:32 PM
|
|
On Sat, 17 Nov 2012, Frederick Williams wrote:
> William Elliot wrote:
> > On Sat, 17 Nov 2012, Daniel J. Greenhoe wrote:
> >
> > > Closure in topological space is defined using at least two different ways in the literature:
> > > 1. cl(A) is the intersection of all closed sets containing A.
> >
> > > 2. cl(A) is the intersection of all neighborhoods containing A, where
> > > a neighborhood is any set containing an open set (an element of the
> > > topology).
> >
> > Those definitions aren't equivalent. Consider Sorgenfrey's two
> > point space S = { 0,1 } with the topology { empty set, {0}, S }.
> >
> > By 1, cl {0} = S while by 2, cl {0} = {0} isn't even a closed set.
> >
> > > Examples of authors who use 1 include Kelley, Munkres, Thron, and McCarty.
> > > Examples of authors who use 2 include Mendelson and Aliprantis & Burkinshaw.
> >
> > > My question is, one definition considered to be more "standard" than the
> > > other (from my very limited survey, 1 might seem more standard).
> >
> > Yes, 1 is the one to be used. 2 is bogus as I showed.
>
> cl(A) 2. should read
>
> cl(A) = {x : for each neighbourhood N of x,
> N intersect A =/= emptyset}.
>
No, that's so much unlike 2, that it can't be a correction of 2.
That's the basic equivalent (as you show below) definition of cl A.
OP is neither confused nor incorrect about 2.
As he indicated, it's a metric space definition and
in fact, in any metric space, 1 and 2 are equivalent.
Are they equivalent in normal T1 spaces?
> A neighbourhood of x is an element of a complete system of
> neighbourhoods of x, denoted N_x. A complete system of neighbourhoods
> of x in X satisfies
>
> For all x in X, N_x =/= emptyset;
> For all x in X and N in N_x, x in N;
> For all x in X and N in N_x, if M superset N then M in N_x;
> For all x in X and N, M in N_x, N intersect M in N_x;
> For all x in X and N in N_x, there is an M in N_x such that M subset N
> and M in N_y for each y in M.
>
> A subset O of X is open if O is a neighbourhood of each x in O. Thus
> "neighbourhood spaces" and topological spaces with the usual open set
> axioms are equivalent.
>
> With those definitions, cl version 1 and cl version 2 are (as one would
> expect) equivalent. I know nothing about these things, but I just don't
> want the OP to be confused.
>
> [neighbourhood = neighborhood]
|
|
