Re: an open set

Andersen <andersen_800@xxxxxxxxxxx> wrote:
Dave Seaman wrote:

That is my problem. Example, X={0,1}, and T={{0}, {1}, {0,1}}. So this
is a topological space as it satisfies (i), (ii), (iii). You have then
declared by fiat that the element {1}, which by definition is a set, is
open. Hence {1} is an open set in this topology?

That's correct. The family of open sets is whatever we say it is,
provided the fundamental requirements (i)-(iii) are satisfied.

Now here is my problem.

Def.1
On the real line, we take an open set to be defined as those sets which
contains at least one non-zero-radius ball centered around every member
of the set. Hence, the interval (1,3) is open. Interval [1,3] is not
open, because there are no non-zero-radius balls centered at 1 that is a
subset of [1,3].

Def.2
Now forget the above and turn to the topological definition. Take X to
be the reals, and T to be the power set of X. It satisfies the 3 rules,
hence it is a topology. the set [1,3] is in the power set of X, hence it
is by definition an open set.

Now Def.1 and Def.2 are contradicting each other, in the first one,
[1,3] is not an open set, in the latter one, [1,3] is an open set.

No, there is no contradiction. The two definition actually perform
different objectives:

Definition 2 sets down the conditions for systems of subsets of a set X
in order to be a topology on X.

Definition 1 specifies a _particular_ system of subsets of the reals,
which fulfills the requirement of Definition 2; so it is a topology
on the reals.

It just happens, that a set (like the reals in this case) may well have
many _different_ topologies.

Marc
.

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