Open Sets & Subspace Topologies
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Thu, 29 Sep 2005 14:24:48 EDT
**This is a homework exercise so only hints please**
If anyone could take a look at my reasoning and let me
know if this seems correct I'd appreciate it.
Consider the subspace Y = [-1, 1] of R where R has the
usual topology T. Then the subspace topology T_y is
defined as the collection {[-1, 1] /\ K : K is in T}.
It's clear to me that it is possible for a set to be
open in Y but not be open in R.
Okay, now let C be the set C = {x : 1/2 <= |x| < 1} which
can be written as C = (-1, -1/2] \/ [1/2, 1). I know
that C is not open in R since C is not an element of T
(that is to say, C cannot be expressed as a union nor as
an intersection of open sets in R). Moreover, C is not
open in Y either, since it is not in the collection T_y.
Now consider the set B = {x : 1/2 < |x| <= 1} which can
expressed as B = [-1, -1/2) \/ (1/2, 1]. This set is not
open in T for the same reasons that C was not open in T.
However, B is open in T_y since the set [-1, -1/2) can
be seen as the intersection [-1, 1] /\ (-n, -1/2) where
n > 1. The argument is similar for (1/2, 1] and thus the
union (which is B) is open in T_y.
These aren't formal proofs, but I just wanted to see if
the reasoning and logic I'm using seems correct.
Thanks in Advance,
Kyle
.
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