Re: connected subsets of R^2
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 8 Sep 2005 17:18:45 GMT
In article <unu0i1hulq46lq7g4m24uv2sgpahtl77r9@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
>Is the following statement true?
>
>If a subset X of R^2 is connected and contains more than one point,
>then X contains a path with more than one point.
No. Consider the set F of all pairs (U,V) where U and V are
disjoint nonempty subsets of R^2, and G the set of all compact
curves C with more than one point. Note that F and G have the
cardinality of the continuum, so well-order F union G by the first
ordinal of that cardinality.
For each (U,V) in F, choose (using Axiom of Choice) a point p(U,V)
that is not in U union V, and for each C in G choose a point p(C)
in C, such that p(U,V) or p(C) is not p(f) for any f < C or (U,V)
respectively in the well-ordering.
Note that R^2 \ (U union V) and C have cardinality c while the set
of points p(f) that must be avoided has cardinality less than c, so
it is possible to do this. Let X be the set of all the p(U,V).
Then X has cardinality C, is connected (for any disjoint nonempty open
sets U,V, X is not contained in U union V because X contains p(U,V)),
but X does not contain any path with more than one point because it
does not contain p(C).
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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