Understanding Bases in Topology: Examining Subsets of Real Numbers

[Damascus Road]

I have a Topology midterm tomorrow and I'm going through exercises in my book. Perhaps someone could let me know whether I ought to make a thread for each question or if I may continue adding to this thread...

Determine which of the following collections of subsets of R are bases:

a.) C1 = {(n,n+2) C R, n $$\in$$ Z }

Two things to show are: That every point in X is contained in a basis element, and that every point in the intersection of two basis elements is is contained in another basis element within that intersection.

So, need to show that every integer is contained in C1. I tend to struggle with the mathematical formulation.. although I can see that obviously an integer n will exist in the set on the real line of n, n+2. How would I start?

 

[Fredrik]

Are you asking if C1 is a basis for the standard topology on ℝ? Use the first condition to show that every open set can be expressed as a union of basis sets. Then consider the set (1/3,2/3).

Btw, try not to use C instead of ⊂. I've seen it before, but it always confuses me. You can copy and paste symbols from http://tlt.its.psu.edu/suggestions/international/bylanguage/mathchart.html , or just make the whole expression a LaTeX expression. Use \subset and remember to refresh and resend after each preview. $A\subset B$.

You should probably just keep adding questions to this thread.

 

[Damascus Road]

I'm assuming the standard topology? I've written the question down exactly as I have it.

Don't I need to first prove condition 1? That every point is in a basis element.

 

[Fredrik]

I've written the question down exactly as I have it.

Then we should probably assume that ℝ has the standard topology.

Don't I need to first prove condition 1? That every point is in a basis element.

I read your post too quickly and misread your condition 1. If you want to show that a given collection of subsets is a basis, one of the things you need to show is indeed that their union is the entire space X. (But your C1 isn't a basis so it would be a waste of time).

What I thought you were saying was this: If X is a topological space, and x is a member of an open set E, then there's a basis set B such that $x\in B\subset E$. (Yes, you didn't say anything like that...I guess it's just what I expected to see). This condition is equivalent to saying that every open set is a union of basis sets. Can you prove that this follows from your definition of "basis"?

My point is that (1/3,2/3) is open in the standard topology, but obviously isn't a union of members of C1.

 

[Fredrik]

So, need to show that every integer is contained in C1. I tend to struggle with the mathematical formulation.. although I can see that obviously an integer n will exist in the set on the real line of n, n+2. How would I start?

You need to know how to show this even though it doesn't help you solve this problem. The solution is actually trivial. Let n be an arbitrary integer. Then n is a member of (n-1,n+1), which by definition of C1 is a member of C1.

Edit: I'll be away from the computer for some time, so someone else will have to help you with the next one.

 

[Damascus Road]

I'm confused why it won't help me solve it. The first requirement for a basis is that every point in X is contained in a basis element.

Thanks for your help!

 

[Fredrik]

Right, but C1 isn't a basis, so if you prove that the first condition is satisfied, that only means that the second isn't.

Edit: The solution I originally had in mind is the one I described above, but we can of course check your condition 2 directly. Is there a basis set that contains 1/2 and is a subset of the intersection of two basis sets? Start by thinking about what basis sets 1/2 is a member of.

I just thought of something else. The method I suggested at first proves that C1 isn't a basis for the standard topology of ℝ. But when we show that the two conditions you listed aren't both satisfied, this means that C1 can't be a basis for any topology on ℝ.

 

[Damascus Road]

Edit: missed something above concerning condition 1. Going over it now. Sorry!

 

[Damascus Road]

Can you explain the choice of n-1,n-1 when my collection is n, n+2?

 

[Fredrik]

Your collection consists of all open intervals (a,b) such that a and b are both integers and b=a+2. (n-1,n+1) is the only such interval that contains n.

$$\{(n,n+2)\subset\mathbb R|n\in\mathbb Z\}=\{(a,b)\subset\mathbb R|a,b\in\mathbb Z, \, b=a+2\}=\{(n-1,n+1)\subset\mathbb R|n\in\mathbb Z\}$$

Uh, wait, why are we proving that every integer is a member of one of these intervals? That's not what condition 1 says. (This time my mistake was that I proved exactly what you asked me to prove. ) Condition 1 is that every real number is a member of such an interval.

You're not supposed to prove that. You can just say that it's "obvious". How you actually prove it depends on what you take as the definition of ℝ. I would define ℝ as an ordered field such that every upper bound has a supremum, and then use the ordered field axioms, but I'm sure that nothing like that is required of you.

 

[Damascus Road]

lol thank you for bearing with me through this.

If condition 1 is, showing that each x in X is in some element of the basis, where C1 is my basis, R is my X, if its not Z?

 

[Fredrik]

These are the definitions and theorems you need to know:Let $(X,\tau)$ be a topological space. The following conditions on a collection $\mathcal B$ of subsets of X are equivalent:

(a) Every open set is a union of members of $\mathcal B$.
(b) If $x\in E\in\tau$, there's a $B\in\mathcal B$ such that $x\in B\subset E$

A collection $\mathcal B$ such that the equivalent conditions above are satisfied is said to be a basis of $\tau$.
Let X be a set, and let $\mathcal B$ be a collection of subsets of X. Suppose that

(1) The union of all the members of B is X.
(2) If $B_1,B_2\in\mathcal B$ and $x\in B_1\cap B_2$, then there's a $B\in\mathcal B$ such that $x\in B\subset B_1\cap B_2$.

Then there's a unique topology $\tau$ on X such that $\mathcal B$ is a basis for it.
If this hadn't been the day before the exam, I would have said that you should definitely prove these claims (the theorems, obviously not the definition) over and over until you can do them with your eyes closed. That might actually still be a good idea. Note that conditions (1) and (2) are the two conditions you stated.

In your specific problem, it's not 100% clear if they're asking

a) Is C1 a basis for the standard topology on ℝ?

or

b) Is C1 a basis for a topology on ℝ?

I think it's the latter. I think that otherwise they should have said "basis for the standard topology" instead of just "basis". It's clear that to verify that condition 1 is satisfied, you would have to prove that every real number is a member of an interval of the form (n,n+2), because ℝ is the set that C1 might define a topology on. (It actually doesn't define a topology on ℝ, but we don't know that until we've given it some thought).

Note that if you've been given a collection B of subsets of ℝ, and it fails to satisfy condition 1, then it might still define a topology on the union of the members of B.

 

[Damascus Road]

Thank you for the very exhaustive help on that!

I'm now looking at interior points, I just want to make sure I understand this properly.

I am considering a topology T on X = {a,b,c,d,e}

T = {X, 0, {a}, {a,b}, {a,c,d}, {a,b,c,d}, {a,b,e} }

And finding the interior points of the subset A = {a,b,c} of X.

My book says, "Note that c is not an interior point of A since c does not belong to any open set contained in A."

But, A is a subset of X, so {a,b,c} is all that is in A, right? But my book says c does not belong to any open set contained in A.

I believe the reason the interior points are a,b are because of the open sets in X: {a}, {a,b}, and the other sets contain elements not in A. I'm having a hard time reconciling the definition above though...

 

[Fredrik]

Last one before I go to bed...

c is an interior point of c if and only if there's an open set E such that $c\in E\subset A$. A set is open if and only if it's a member of the topology. That's just the definition of "open". So you have specified that the only open sets are:

∅
X
{a}
{a,b}
{a,c,d} <---------
{a,b,c,d} <---------
{a,b,e}

The arrows mark the only two that contain c. Neither of them is a subset of A. Therefore, there is no open set E such that $c\in E\subset A$, and that means that c is not an interior point of A.

If you need to ask more questions tonight, you will probably get an answer more quickly if you ask it in the homework forum.

 

[Fredrik]

So how did the exam go?