Re: compact groups and AC
- From: "Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx>
- Date: Sun, 24 Jul 2005 17:02:54 -0400
Stephen J. Herschkorn wrote:
Does the proof of the following need the axiom of choice?
If a T1 topological group G has a compact subgroup H such that G/H is compact, then G is compact.
In case anyone is interested, the answer is no. Using the same technique I demonstrated elsewhere in this thread, I removed choice from my proof of the following:
A *perfect* map is by definition a closed, continuous surjection each of whose fibers is compact. A topological space is compact if and only if there exists a perfect map thence to a compact space.
Proof of the "if" part:
Notational comment to ease reading: Since we can't use script font in these posts, lower-case v and w will represent open sets, and upper-case U, V, and W will represent collections of open sets.
Let f: X --> Y be perfect, Y be compact, and U be an open cover of X. For each y in Y, let
V_y = {union(V): V is a finite subset of U and V covers f^(-1) (y)}. That is, v in V_y iff
f^(-1) (y) is a subset of v and v is the union of a finite subcollection of U
For each y in Y, let W_y = {Y \ f(X \ v): v in V_y}; note that for all y in Y, y in w for all
w in W_y. Let W = union(y in Y, W_y). W is an open cover of Y, hence there exist
w1,..., w_n in W which cover Y. Let y_i in Y be such that w_i in W_(y_i), and let
v_i in V_(y_i) be such that w_i = Y \ f(X \ v_i). Then {v1,..., v_n} covers X.
q.e.d.
-- Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx Math Tutor in Central New Jersey and Manhattan
.
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- From: Stephen J. Herschkorn
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