Re: Continuous maps and extending into closure
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Wed, 1 Nov 2006 17:06:56 +0000 (UTC)
In article <1162400198.682463.258770@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Jason Pawloski <jpawloski@xxxxxxxxx> wrote:
If f:X->X is a continuous function on some metric space, and for some
subset of X denoted K, f(K) is a subset of K, how do I show that f(K~)
is a subset of K~ where K~ is the closure of K.
If you don't mind, I'll use cl(K) instead of K~. It's just so I won't
make a mistake...
Proving f(cl(K)) is contained in cl(K) is equivalent to proving that
X\cl(K) is contained in X\f(cl(K)). So let x be in the complement of
K~; we want to show that x is in the complement of f(cl(K)).
If x is not in f(X), then we are done. So assume that there exists y
such that f(y) = x.
Since cl(K) is closed, X\cl(K) is closed. So there exists e>0 such
that B(x,e), the open ball with center in x and radius e, is contained
in X\cl(K).
Since f is continuous, and f(y)=x, there exists a d>0 such that
f(B(y,d)) is contained in B(x,e). In particular, f(B(y,d))/\cl(K) is
empty. Since f(K) is contained in K, that means that B(y,d)/\K is
empty (if z were in B(y,d)/\K, then f(z) would be both in B(x,e) and
in f(K), which is contained in K, which is contained in cl(K)).
Therefore, there is a neighborhood of y which is disjoint from K, so y
is not in cl(K). Therefore, f(y) is not in f(cl(K)). But since f(y)=x,
that means that x is not in f(cl(K)), which is what we wanted to
prove.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org
.
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