Re: Co-countable topology
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Mon, 29 Aug 2005 17:21:59 +0000 (UTC)
In article <deverb$2p6n$1@xxxxxxxxxxxxxxxxxx>,
Arturo Magidin <magidin@xxxxxxxxxxxxxxxxx> wrote:
>In article <Y1HQe.116052$fm.7646850@xxxxxxxxxxxx>,
>computer <computer@xxxxxxxxxxxxxxxxx> wrote:
>>Consider a sequence {x_n} in R with the co-countable topology. Does {x_n}
>>converge if it is constant?
[...]
>Did you mean to ask whether the sequence converges if and only if it
>is (eventually) constant?
>
>In this case, yes. Suppose that for every N there exists i,j>N such
>that a_i is different from a_j. Let x be any element of R. The
>set S = {x_n : n in N} - {x} is nonempty and countable, so (R-S)U{x} is a
>neighborhood of x. Since for every N there exists j>N such that x_j is
>different from x, it follows that the sequence is not eventually in
>(R-S)U{x}, so the sequence cannot converge to x.
Hmmm... A direct proof would be better than this. Basically the same
argument. Let S = {x_n : n in N}, and assume the sequence converges to
c. Then (R-S)U{c} is a neighborhood of c, hence there exists N such
that x_i in (R-S)U{c} for all i>N. But S intersect (R-S)U{c} is
contained in {c}, hence x_i in {c} for all i>N; therefore, x_i=c for
all i>N, showing the sequence is eventually constant.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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