What is the definition of a topological space?

[JasonRox]

This is simply a self-study.

1 - Can someone help understand what they mean by the following Topological Space:

Let T be the family of subsets of R (the reals) defined by: A subset K or R belongs to T if and only if...

...for each r in K there are real numbers a, b such that...

a < r < b and {x: xeR, a < x < b} C K.

Note: C was used to show that it is a subset of K.

Are they trying to say that K is finite?

2 - For the Discrete Topology for X, which subsets are closed?

My answer is that they are all closed and open because all the subsets are included, hence they are all open, but by definition the complement of each of those sets is closed. Since the complement the those sets are also possible subsets of X, they are all open since the Discrete Topology contains all the subsets.

3 - For the Indiscrete Topology for X, which subsets are closed?

My answer is all the subsets of X are closed. The Indiscrete Topology for X only contains the void set and the underlying set X, which both of which are also closed, as well as every subset of X.

I'd appreciate any comments.

 

[DeadWolfe]

1 Not any finite subset is in T. For instance [0,1] is not in T. (0,1) is however. Why?

2 - You are right here.

3 - Not every subset is closed. Consider R with the indiscrete topology. (0,1) is not closed since R/(0,1) is not equal to R or the null set, and hence is not open. Given a set X with the indiscrete topology, to find the closed sets, find the sets whose complements are X and the null set. But that shouldn't be too hard, right?

 

[JasonRox]

1 Not any finite subset is in T. For instance [0,1] is not in T. (0,1) is however. Why?

2 - You are right here.

3 - Not every subset is closed. Consider R with the indiscrete topology. (0,1) is not closed since R/(0,1) is not equal to R or the null set, and hence is not open. Given a set X with the indiscrete topology, to find the closed sets, find the sets whose complements are X and the null set. But that shouldn't be too hard, right?

1 - I know the difference of [0,1] and (0,1).

Let T be the family of subsets of R (the reals) defined by: A subset K or R belongs to T if and only if...

...for each r in K there are real numbers a, b such that...

a < r < b and {x: xeR, a < x < b} C K.

This seems like [0,1] can be in T if you make a and b, like (-1,2). It asks for a and b such that all the elements in K are between a and b. As long as K is finite, you can find such elements.

3 - Ok, I know that the void set and the underlying set X is open. I was just saying that all the subsets of X are closed, but the void and X set are also open.

Thanks, for responding. I appreciate it.

 

[matt grime]

1 - I know the difference of [0,1] and (0,1).
This seems like [0,1] can be in T if you make a and b, like (-1,2). It asks for a and b such that all the elements in K are between a and b. As long as K is finite, you can find such elements.

Nope, you are asserting that the open interval is (-1,2) is contained in [0,1].

The definition is U is open if for any point p in U there is an open interval containing U and strictly contained in U. ( that is, you have the definition exactly reversed: you should read all points between a and b are in K NOT all points in K are between a and b)

3 - Ok, I know that the void set and the underlying set X is open. I was just saying that all the subsets of X are closed, but the void and X set are also open.
Thanks, for responding. I appreciate it.

No, again, a set is closed if and only if its complement is open, since there are exactly 2 open sets there are two closed sets, that is if, as I read it, you're claiming all sets are closed in the indiscrete topology.

 

[JasonRox]

Nope, you are asserting that the open interval is (-1,2) is contained in [0,1].

The definition is U is open if for any point p in U there is an open interval containing U and strictly contained in U. ( that is, you have the definition exactly reversed: you should read all points between a and b are in K NOT all points in K are between a and b)

No, again, a set is closed if and only if its complement is open, since there are exactly 2 open sets there are two closed sets, that is if, as I read it, you're claiming all sets are closed in the indiscrete topology.

1 - I get it now. If K contains all elements [0,1], you can't find a and b such that it will satisfy the definition. You can make the number as close as you'd like to 0 from the negative side, but there is always going to be a real number that is closer that K does not contain.

3 - Oh, I see what's going on here. There is only two sets, so I can only speak of the sets that are in T, and the complements of the sets in T. Right?

Thanks again.

 

[matt grime]

I made a typo, it shuold read:The definition is U is open if for any point p in U there is an open interval containing ******p**** and strictly contained in U.

 

[JasonRox]

I made a typo, it shuold read:


The definition is U is open if for any point p in U there is an open interval containing ******p**** and strictly contained in U.

I'm really confused now.

I got this part...

The definition is U is open if for any point p in U there is an open interval containing ******p****

...so I got that, but I have no clue what you are talking about here...

and strictly contained in U.

So, what is strictly contained in U? The interval? The element P?

 

[matt grime]

The interval, obviously.

Once more: in topology 1. A set U is open if for all p in U there is an open interval $$I \subset U$$ such that $$p \in I$$.

It is just the usual (metric) topology on R where open sets are (all) unions and finite intersections of open intervals, and all closed sets are the finite union and arbitrary intersection of closed intervals, which is exactly the metric topology: a set U is open iff there exists an e such that open ball B(x,e):={y: |x-y| <e} lies in U. Given that I stated that p was a point in U, that point is obviously (strictly) contained in U.

 

[JasonRox]

The interval, obviously.

Once more: in topology 1. A set U is open if for all p in U there is an open interval $$I \subset U$$ such that $$p \in I$$.

It is just the usual (metric) topology on R where open sets are (all) unions and finite intersections of open intervals, and all closed sets are the finite union and arbitrary intersection of closed intervals, which is exactly the metric topology: a set U is open iff there exists an e such that open ball B(x,e):={y: |x-y| <e} lies in U.


Given that I stated that p was a point in U, that point is obviously (strictly) contained in U.

Thanks, for clearing that up.