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Topic: Connected Subspaces
Replies: 2   Last Post: Jun 30, 2014 10:27 AM

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 William Elliot Posts: 1,556 Registered: 1/8/12
Re: Connected Subspaces
Posted: Jun 30, 2014 10:27 AM

In article <290620141155554340%edgar@math.ohio-state.edu.invalid>,
Peter Flor <peter.flor@chello.at> wrote:

> The answer is contained essentially in the topology textbook by K.
> Kuratowski.
> Let M be a space as required: metric, connected, card(M)>1. Then M is
> at least of continuum cardinality:
> if 0<t<d(a,b) for some a and b in M then the set of all x such that
> d(a,x) = t cannot be empty since M is connected. (If empty it would
> furnish an obvious decomposition of M.) Thus we have continuum many
> nonempty, pairwise disjoint sets in M, and a fortiori, continuum many
> points in M.

Cute proof.  Another proof.  Since M is completely Hausdorff, for
distince
a,b in M, some Urysohn function f:M -> [0,1] with f(a) = 0, f(b) = 1.
Since M is connected, f(M) = [0,1].  Thus c <= M.

> Next, let M_1 be the set of those x{\in}M disconnecting M, and M_2 its
> complement.  Then at least one of M_1 and M_2   has the same
> cardinality as M. All sets M\{x},  x{\in}M_2 , are connected, and they
> are pairwise distinct. So if card(M_2) = card(M) we are done. Moreover
> by a certain lemma (Thm. 4 in  Kuratowski, Chapter 46 II), if x{\in}M_1
>  and A{\cup}B is a decomposition of M\{x}, then both A{\cup}{x} and
> B{\cup}{x} are connected. Obviously, couples [A{\cup}{x}, B{\cup}{x}]
>  are different for different values of x. Hence, the set of these
> couples has the same cardinality as M_1. In either case, the set of
> connected subsets of M has uncountable cardinality.

Additionally |A \/ {x}| or |B \/ {x}| = |M|.

Can you show for each point p, that p in in many |M| connected subsets.
How many of those subsets would have cardinality |M|?

Date Subject Author
6/25/14 William Elliot
6/29/14 Peter Flor
6/30/14 William Elliot