# Discrete Topology on RWatch

#1

I would like to see how to start this.

Definition (Discrete topology): Let be a set, and be the collection of all subsets of . Then is the discrete topology on . In other words, is discrete every is open.

So as I understand it, I need to show that every interval with is open? Even though it says that they are all open in the question?
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5 days ago
#2
(Original post by RDKGames)

I would like to see how to start this.

Definition (Discrete topology): Let be a set, and be the collection of all subsets of . Then is the discrete topology on . In other words, is discrete every is open.

So as I understand it, I need to show that every interval with is open? Even though it says that they are all open in the question?
I'm not sure where your last paragraph has come from. You need to show that every subset of R is open under this topology. (e.g. the set of all irrational numbers is open, the set {sqrt{2}} is open, etc...)

Hint:
Spoiler:
Show
Obviously if every subset is upon, you're going to need to show every point set {x} (for some x in R) is open. This is straightforward to show from the information given and one of the definitions of a topology. And then use another definition to finish.
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#3
(Original post by DFranklin)
I'm not sure where your last paragraph has come from. You need to show that every subset of R is open under this topology. (e.g. the set of all irrational numbers is open, the set {sqrt{2}} is open, etc...)

Hint:Obviously if every subset is upon, you're going to need to show every point set {x} (for some x in R) is open. This is straightforward to show from the information given and one of the definitions of a topology. And then use another definition to finish.
I think I understand the gist of the argument. If we are able to show that every point in is open (i.e. an element of ), then is comprised of every possible union of these elements (due to definition), and so contains every possible subset of , hence we have discrete topology?

Also I think the terminology is throwing me off a bit. At first I was confused since square brackets on mean the set is closed, but in this context 'open' just refers to the closed set being an element of , right?
Last edited by RDKGames; 4 days ago
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4 days ago
#4
(Original post by RDKGames)
I think I understand the gist of the argument. If we are able to show that every point in is open (i.e. an element of ), then is comprised of every possible union of these elements (due to definition), and so contains every possible subset of , hence we have discrete topology?

Also I think the terminology is throwing me off a bit. At first I was confused since square brackets on mean the set is closed, but in this context 'open' just refers to the closed set being an element of , right?
Yes
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#5
(Original post by DFranklin)
Yes
Ah thanks.

Is this a good enough argument to show that every point in is an element of ?

Since for , then it's true that where . Then since is a topology, it must mean that .
We can extend this to say that for all . Similarly, .

So every point point on the real line must be in .
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4 days ago
#6
(Original post by RDKGames)
Ah thanks.

Is this a good enough argument to show that every point in is an element of ?

Since for , then it's true that where . Then since is a topology, it must mean that .
We can extend this to say that for all . Similarly, .

So every point point on the real line must be in .
I don't really understand what/why you're doing to "extend" your first result.

Why wouldn't you just do something along the lines of:

Claim: {x} \in T for all x in R. Proof: pick x; we know [a, b] \in T if a<b. So [x-1,x] and [x, x+1] \in T. So their intersection is in T, so {x} \in T.

[If it's easy to be "concrete" about how to construct something, it usually makes explaining your construction easier (particularly justifying that it actually works)].
1
#7
(Original post by DFranklin)
I don't really understand what/why you're doing to "extend" your first result.
I am shifting the original intervals to the right by and showing that the number is in for any . To me at least this is the same as saying that every possible number greater than is in . Then I repeat this argument to get that every possible number less than is in .

Why wouldn't you just do something along the lines of:

Claim: {x} \in T for all x in R. Proof: pick x; we know [a, b] \in T if a<b. So [x-1,x] and [x, x+1] \in T. So their intersection is in T, so {x} \in T.

[If it's easy to be "concrete" about how to construct something, it usually makes explaining your construction easier (particularly justifying that it actually works)].
I was trying to think of a simpler way but I couldn't think of anything better than what I had ended up with. I'll note this approach though alongside my own if its valid
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4 days ago
#8
(Original post by RDKGames)

I would like to see how to start this.

Definition (Discrete topology): Let be a set, and be the collection of all subsets of . Then is the discrete topology on . In other words, is discrete every is open.

So as I understand it, I need to show that every interval with is open? Even though it says that they are all open in the question?
These intervals are taken to be open as they're in the topology. To show that the topology is the discrete topology you need to show that every set in R is open, which should be quite easy considering the union [a,p] n [p, b] is open.

In fact it can be shown that every topology with the singleton set open is discrete, once you've done this question the proof of this statement will be trivial.
Last edited by Ryanzmw; 4 days ago
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4 days ago
#9
(Original post by RDKGames)
I am shifting the original intervals to the right by and showing that the number is in for any . To me at least this is the same as saying that every possible number greater than is in . Then I repeat this argument to get that every possible number less than is in .
So, on one level, It's just weird to say "we can extend this", and then basically just restate what you've already said.
And then the fact that you talk about "epsilon > 0" (which does have a traditional meaning), and talk about shifting intervals (and seem to think you need another argument to shift to the left) is concerning. In questions like this (about general topologies), you need to be extremely careful about using "common sense notions" about how behaves. In this particular case, any kind of "epsilon-delta" thinking is wrong - the topology behaves entirely differently. And there's not really any meaningful idea of being able to "shift" things either.

I was trying to think of a simpler way but I couldn't think of anything better than what I had ended up with. I'll note this approach though alongside my own if its valid
I'm not wanting to flog a dead horse here; we've all written super-ugly proofs that might manage to be "valid", but miss the point and take 3 times longer than they should because it was the method that came to mind. At the same time, in this case, I think you should try to understand why your method is basically just proving essentially the same result 3 times for no good reason.
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