Topology question

OP

hmmm. i haven't done any metric spaces so that was my mate that was wrong with that defn.
so we're just saying that openness is some property possesed by elements of the topology.

now, i'm trying to show

\mathfrak{U}

is a dsicrete topology, that is that every possible subset of

\mathbb{Z}

is open i.e. in

\mathfrak{U}

. so i need to show that every possible subset is open and to do that i need to show that {x} is open for all x in Z. correct?

well, if all infinite subsets are open then surely we can write a particular infinite subset as a union of sets {x}. e.g. Z = infinite union of {x} for all x in Z.
would that force each {x} to be open or have i screwed up again?

Reply 22

generalebriety

16

latentcorpse

i need to show that every possible subset is open and to do that i need to show that {x} is open for all x in Z. correct?

Yes.

latentcorpse

well, if all infinite subsets are open then surely we can write a particular infinite subset as a union of sets {x}. e.g. Z = infinite union of {x} for all x in Z.
would that force each {x} to be open or have i screwed up again?

No. You've shown that a union of sets {x} is open, but there is no rule saying that this means {x} is open too. I refer you back to my example of the topology containing only the empty set and Z; then {x} is not open for any x in Z, but their union is (because their union is Z).

Reply 23

IrrationalNumber

I'm wondering if you're confused over the difference between a topology and a set of points, they're different things. A topology on a set of points is a group of subsets from the set of points (not necessarily all subsets) whereas the set of points is just a collection of numbers (for example

\mathbb{N}

might be a set of points, but {

\mathbb{N}

,

\emptyset

} might be a topology on the set of points)

Reply 24

OP

ok, can i get a hint about what to take the intersection of in order to prove {x} is open because i really have no idea. sorry.

Reply 25

ghostwalker

17

A somewhat simplistic description, but:

Stand on the number line and look to infinity, and what do you see (including yourself).

Spoiler

Reply 26

OP

aahhhh!

so

(-\infty,x]

and

[x,\infty)

are both infinite and hence open

\forall x \in \mathbb{Z}

and so by the properties of a topology

(-\infty,x] \cap [x,\infty)=\{ x \}

is open

\forall x \in \mathbb{Z}

.

That tells us

\{ x \} \in \mathfrak{U} \forall x \in \mathbb{Z}

and since every possible subset of

\mathbb{Z}

can be made by taking unions of

\{ \{ x \} | x \in \mathbb{Z} \}

we conclude that

Unparseable latex formula:

\mathfrak{U} = \powerset ( \mathbb{Z})

and hence

\mathfrak{U}

is the discrete topology.

i wanted it to write U= power set of Z but couldnt get it to work in latex.

is that ok? is there are more concise way of putting it?

Reply 27

Yes, that's fine. Not really any better way of doing it.

Reply 28

OP

ok thanks. i don't see how for the second part of the question i need to use the first part that i just showed though.
why can't it be done exactly the same way. pick a random point and take the intersection of two half-open sets etc...

Reply 29

Because a random point is not the same as a random subset.

Reply 30

OP

not sure i follow.

what if

X=(-\infty,...,a,b,c,d,e,f,g,...,\infty)

why can't i just take

(-\infty,c]

and

[c,\infty)

as two infinite subsets and then intersect them as before?

Reply 31

What if X = {1,3,5,7}

What two infinite subsets are you going to intersect to give X?

Reply 32

OP

but i thought X was infinite. {1,3,5,7} isn't infinite?

Reply 33

If you want to prove the topology is the discrete topology then you need to prove that every subset of Z is open in the topology. {1,3,5,7} is certainly a subset of Z.

Reply 34

ghostwalker

17

latentcorpse

ok thanks. i don't see how for the second part of the question i need to use the first part that i just showed though.
why can't it be done exactly the same way. pick a random point and take the intersection of two half-open sets etc...

DFranklin

Because a random point is not the same as a random subset.

Correct me if I'm wrong, particularly as your knowledge of topology is considerably more extensive than mine, but I think you may have gone off track.

In the first part, considering the integers, latentcorpse has shown that the subset of any single point is open in the topology; and hence any subset is open, simply by taking the union of single point sets; and hence we have the discrete topology.

This is what he is trying to do with the second part of the question, for a general infinite set. His problem is, once again, to show that a single point subset is open, and once this is achieved, the rest will follow. The achievement of that first bit however, not being quite so simple.

Reply 35

OP

ok yeah i follow what ghostwalker said - i thought that's what i was trying to show also.

i'm assuming my attempt earlier was wrong as it didn't involve using the first part of the question.

any hints here. is it some sort of map from the integers to another infinite set?

Reply 36

generalebriety

16

latentcorpse

not sure i follow.

what if

X=(-\infty,...,a,b,c,d,e,f,g,...,\infty)

why can't i just take

(-\infty,c]

and

[c,\infty)

as two infinite subsets and then intersect them as before?

Because we're no longer necessarily working in the set of integers. We might be working in R, or C, or S², or something much less familiar.

Reply 37

generalebriety

16

Incidentally, I think it's only really the method of the first part that's meant to be used, i.e. taking an intersection to show that {x} is open in X. This involves finding two disjoint infinite sets in X - something I don't think is very easy to do (though maybe I'm being stupid).

Reply 38

IrrationalNumber