Re: Countable basis implies separability?

Michael Stemper wrote:
In article <Gz6sh.68487$wP1.2917@xxxxxxxxxxxxxxxxxxxxxxxxxx>, W. Dale Hall writes:
Michael Stemper wrote:
In article <QYarh.20541$yC5.14574@xxxxxxxxxxxxxxxxxxxxxxxxxx>, W. Dale Hall writes:
Michael Stemper wrote:

I'm trying to prove that any open set with a countable open basis
is separable. To do this, I need to show a countable dense
subset. (Unless I've completely misunderstood the material.)

I'm guessing that I somehow generate this subset by selecting one
(or possibly countably many) points from each set in the basis,
and then showing that the subset is dense.

Hint: *every* open set contains (as a subset) at least one of the
elements of any basis of the topology.

Unfortunately, topological spaces aren't introduced until section
4.5, and this is the last problem in 4.3.

Wow. If I didn't know what a topological space is, then
I wouldn't know how to ask your question.

Well, I feel less stupid then. I've been mulling this one over since
before Christmas with no progress. Twas a mite discouraging.

Let's back up then.

Here's the question:

I'm trying to prove that any open set with a
countable open basis is separable. To do this,
I need to show a countable dense subset.

Now, let's do a little vocabulary. Please give your
definitions for the following terms.

OPEN SET

In a metric space, (E,d), X is an open set if, for all x in X, there exists
an r(x) such that B(x,r(x)) is a subset of X.

(An open ball, B(p,q) is the set of all points s such that d(p,s) < q.)

BASIS

A collection of open sets, C, is an open basis for an open set X if,
for any open set, Y in X, Y is equal to the union of members of C.

SEPARABLE

A set K is separable if it has a countable dense subset.

DENSE

A set G is dense in H if and only if the closure of G is H.

(The closure of G is the union of int(B) and bdy(G).)

Lemma. x is in the closure of G iff every open set containing x
intersects G.

Proof assuming the lemma: Let S be the open set. Let C = {C_i} be the
countable open base for S. Let D = {d_i} where d_i is in C_i. We want
to prove the closure of D is S. Use the lemma: Let x be in S. Let Y be
a subset of S that is open and contains x. Let J = {i| C_i subset of
Y}. Then Y = U_{i in J} C_i because C is a base. Since x is in Y, J is
not empty. Let j be in J. D contains a point in C_j. So, C_j
intersects D. Hence, Y intersects D.

--
David Marcus
.

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