Dense sets

Hello all, can someone explain to me why the definition in the picture is so?

The definition of a dense set is that given a proper subset A of X, its closure is the entirety of X.

My assumptions:
A closure cannot add open parts of X to A, since the 'objective' of a closure is to make the complement of A open
i.e. the objective of a closure is to take all the boundary points of A and include them in A.
That means that for A closure to equal X, A must already have all the open subsets of X, namely the interior and exterior of A, and also possibly part of the boundary. But the only way A can contain its exterior is if its exterior is null!
my conclusion: the exterior of a dense subset is null. But then how is A a proper subset of X??

Now look at the attached picture. It says that A is dense in X iff all nonempty open subsets U of X are such that U intersection A is not null.
But if A contains all the 'open' part of X, surely any open subset of X is contained in A, not just intersecting?? For an open subset of X to be intersecting A but not contained in it means that it is a union of a subset of Int(A) with a subset of Ext(A). But as I previously wrote, it doesnt make sense for Ext(A) to not be null!!

Can someone please clear up my confusion? I hope my question wasn't too convoluted.. I will try and explain any unclear points.

It means everything is close to the set
duh

I can't see the contradiction from this. Can you give me an example in (X,Tp) ?

what does "everything is close to the set" mean?

If A is a proper subset of X, then the density of A in X means that the closure of A in X is X. If we then take any open set U in X, suppose its intersection with A is empty. There is then some x in X-A such that it has a nbd not meeting A, so x is not contained in the closure of A, and thus the closure is not X and A is not dense in X.

>then some x in X-A such that it has a nbd not meeting A, so x is not contained in the closure of A, and thus the closure is not X and A is not dense in X

Isn't this the same as saying that the nbd of x is completely contained in Ext(A)?

But any open neighborhood of a point outside of A has to be completely contained in Ext(A), otherwise it would cross the boundary of A and not be open anymore. Since neither of these can exist when A is dense, doesn't that mean that Ext(A) = {} ?

Indeed, it is the same. For any subspace A of a topological space X, X can be decomposed into three disjoint subsets, the interior, the exterior and the boundary of A. The closure of A is the union of the interior and the boundary of A, and to be dense is to have an empty exterior.

>A contains all the 'open' part of X

No.
Q is dense in R (by construction)
Q∪{√2} is dense in R too

>surely any open subset of X is contained in A

(1,2)⊈Q since √2∉Q

Ok so Ext(A) = {} therefore any open subset in X has to be completely contained in Int(A), which means that any nonempty open subset of X is a subset of A, correct?

Then why did the professor define it as in the attached picture?

No, it is not required that that every open set is included in A. The rationals are dense in reals under the ordinary topology of R, but Q contains no open sets. The interior of Q is empty, but so is the interior of R-Q.

this

*no nonempty open sets

I think I'm beginning to understand, √2 is in Ext(Q) but is infinitely close to points in Int(Q). In this case, what is the boundary of Q? Does it exist?

I can see why R-Q has no interior, as it is just isolated irrational points, but how is Int(Q) also empty, since you can make an open subset say from (1,√2)

(1,√2) is an open set that is completely contained in Q.. or are there infinite irrational gaps within (1,√2)?

Yes, ofc.

Every open interval contains both a rational and an irrational number. Since the ordinary topology in R is generated by open intervals, each open set in R contains both or is empty. For an open set to be contained in Q, it must therefore be empty.

Alright, understood, thanks guys

No problem, m808

Since this thread is open I may as well ask something else.

Here the prof is proving that whenever B1 and B2 exist in the basis, B1 int B2 must also exist, however I have come up with an example to the contrary that doesnt seem to break any axioms, namely that while U1 int U2 is in the topology Tau, it can be constructed from B3 alone and doesnt require there to be a B4 = {x} which is a subset of B1 int B2.

W-what? Could you write it a bit more clearly? I can't grasp your idea.

Times like this OP is where it's good to take a step back and just write out the definition in English, maybe draw some pictures to feed your intuition. Don't be afraid to do that. Any intuition you can get about Topology is very nice.

This. It makes things really clear in the beginning, and thus functions as a stepping stone into the realm of more complicated stuff. For example, the result that the union of overlapping connected sets is connected is obvious after one visualizes it.

This just isn't true. Open balls [math]B_\epsilon(v)[/math] are a basis for the standard topology on R^n but the intersection of two balls isn't a ball. Maybe your professor just meant there was a basis element contained in it. That's one way to define a basis.

For a subset [math]\mathcal{B}[/math] of a topology to be its basis, it is required that for any [math]B_1, B_2 \in \mathcal{B}[/math] with [math]B_1 \cap B_2 \neq \emptyset[/math], there exists [math]B_3 \in \mathcal{B}[/math] such that [math]B_3 \subset B_1 \cap B_2[/math]. Just like said.

I hope I didn't fuck up the latex. My fingers are constantly on wrong keys today.

Ah I see now, it was B3 exists iff its a basis of the topology, not 'B3 always exists'. Thanks doc

Also I didn't realize you could type Latex here. Pretty cool!

No problem, m808!

How to do it is left as an excercise.

>what does "everything is close to the set" mean?
Open sets are supposed to suggest nearness. So no matter how close you look to any point (i.e. any open neighborhood of any point), you are going to find points of your dense set.

Everything : the whole space X
is close to: is included in the closure
the set: A

[math]X \subseteq \bar{A} [/math]

>since this thread is open
>open
HAHAHAHAHAHAHAHA

After consolidating the view of Q being dense in R, I now know where you were going with this.. rational numbers infinitely close to irrational numbers no matter how 'zoomed in' you are. Thanks

Exactly. This notion of something happening near any point is a big idea that comes up a lot as you continue on. It's a very good idea to continue to think of things in terms of R at first, since this is where many notions come from.