# Proving T is a topology.

## Homework Statement

Let X be a set and p is in X, show the collection T, consisting of the empty set and all the subsets of X containing p is a topology on X.

## Homework Equations

?
A topology T on X is a collection of subsets of X.
i) X is open
ii) the intersection of finitely many open sets is open
iii) the union of any collection of open sets is an open set.

## The Attempt at a Solution

How do I know that X is open. Maybe open doesn't mean the same thing as an open set on the reals.
if X is open. And If I take an intersection of a finite number of sets in T i will get something in T so by definition It will produce an open subset in T. And all my intersections will contain P because P is in every set.
for iii) If I take a union of all the subsets in T I will produce a subset in T containing p. I could not produce something outside of T because i started with sets that were part of
T.

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Bacle2
Does X contain p? This is the way of telling if X is open in your topology.

so because we know that p is in X , we know that it is open.
Is a topology just a way of defining properties on our weird set.

Bacle2
Well, after a lot of time , it was found that many of the properties that were

usually studied in topology could be defined in terms of open sets: continuity,

compactness,etc. And, yes, by assigning a topology to a set ,we gain the ability

to talk about topological concepts in the set. And with different topologies, we

end up with different topological properties.

HallsofIvy
Homework Helper
A topology T on X is a collection of subsets of X.
i) X is open
ii) the intersection of finitely many open sets is open
iii) the union of any collection of open sets is an open set.
Are you sure you have copied this correctly? You seem to be defining a "topology T" but then don't mention "T" in the definition! But you do talk about "open sets" without defining them.

What is your definition of "open" set? Normally, in General Topology, we define a topology, T, of X as a collection of subsets of X satisfying:
1) X is in T
2) The empty set is in T
3) The union of any subcollection in the T
4) The intersection of any finite subcollection is in the T

We then define a set to be "open" if and only if it is in the collection.

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Bacle2
Are you sure you have copied this correctly? You seem to be defining a "topology T" but then don't mention "T" in the definition! But you do talk about "open sets" without defining them.

What is your definition of "open" set? Normally, in General Topology, we define a topology, T, of X as a collection of subsets of X satisfying:
1) X is in T
2) The empty set is in T
3) The union of any subcollection in the T
4) The intersection of any finite subcollection is in the T

We then define a set to be "open" if and only if it is in the collection.
I thought we define a set U to be open if for every x in U there is an element Ux of T

as above, such that x is contained in Ux , and Ux is in T.

ok thanks for your help guys

vela
Staff Emeritus
Homework Helper
I thought we define a set U to be open if for every x in U there is an element Ux of T

as above, such that x is contained in Ux , and Ux is in T.
What do you mean by Ux?

It's been awhile since I took topology, but I remember it the way HallsofIvy said. If a set is in T, it is, by definition, open.

SammyS
Staff Emeritus
Homework Helper
Gold Member
What do you mean by Ux?

It's been awhile since I took topology, but I remember it the way HallsofIvy said. If a set is in T, it is, by definition, open.
Yes vela & Halls are correct. In fact the topology described in this problem is called the particular point topology (or included point topology).

Yes vela & Halls are correct. In fact the topology described in this problem is called the particular point topology (or included point topology).
The particular point topology is the answer to an interesting question.

We know that if K is a compact topological space, every continuous function from K to the reals is bounded.

Now, if X is a topological space such that every continuous function from X to the reals is bounded, must X be compact?

The answer is (surprisingly) no. The counterexample is any infinite set with the particular point topology. The proof is simple and amusing.

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