Topology: Proving Z Open in X/Y iff Y ∪ Z Open in X

[latentcorpse]

Given a topological space X and a subspace Y \subseteq X let X / Y be the union of the complement X \backslash Y and a set with one point. Define an equivalence relation \tilde on X such that X/~=X/Y, and use it to deifne a topology on X / Y as an identification space of X, with projection p: X \rightarrow X / Y. Prove that a subset Z \subseteq X is such that p(Z) is open in X / Y if and only if Y \cup Z is open in X.

ok. so i thought the equivalence relation should identify points outside Y wiht points inside Y as that way the quotient would be equal to the complement. but i want the quotient to be equal to X/Y and because its the union of the copmlement and this singleton, I'm confused - this extra one point set is making it hard to see a relation.

 

[Billy Bob]

Does it work to send each element in X minus Y to itself, but everything in X intersect Y to the single point? That is, collapse X intersect Y to one point.

 

[latentcorpse]

yeah i think i have

x~y if (x=y) or (x \in Y, y \in Y)

that forces X/~ = X/Y

what about the next bit with defining the topology and the projection though?

 

[Billy Bob]

Let t denote the single, distinguished point.

p:X -> X/Y is p(x)=x if x is not in Y, and p(x)=t if x is in Y.

I think the above is easier to understand. If you want the range of p to be the set of equivalence classes instead, I guess you could define

p(x)={x} if x is not in Y, and p(x)=X intersection Y if x is in Y.

To define the topology, use your definition of identification space or quotient space or quotient topology.

 

[latentcorpse]

hi. thanks for the reply. i have a couple of questions:

(i) you wrote out two posisble functions p. are both correct? the first one makes sense but i don't understand the second one.
our definition of p is that it is the canonical surjection that assigns each x \in X the equivalence class [x] \in X/~
i.e. p:X \rightarrow X/~ ; x \rightarrow [x]
surely i would then want p to have its' range as the set of equivalence classes in order to agree with my definition? but why does p map to X intersect Y for x in Y? X intersect Y=Y and this is a set of many points i thought the whole point was that everything in Y was made equivalent to one point only?

(ii)Our deifnition of an identification space/quotient space is that it's the topological space consisting of the equivalence classes such that a subset U \subseteq X/~ is open if and only if p^{-1}(U) \in X is open.

i don't really understand how to use this to define a topology?
i thought to define a topology all we need is to say what the topological space is, in this case X/~ and the set of all open sets \mathfrak{U} and then you can just say the topology is (X/~,\mathfrak{U}).
i'm still very new to this and don't really understand what's going on but I am 99% sure the above is wrong because i haven't actually done anything!
what is it I'm required to do? work out what all the possible open sets are?

 

[Billy Bob]

(First of all, I forgot that Y was a subset of X. That's why I kept saying "X intersect Y" instead of simply "Y.")

Let's just use the first definition of p, since it makes sense to you.

p:X->X/Y given by p(x)=x if x not in Y, p(x)=t if x in Y.

Example: if X=[0,10] and Y=[2,4]\cup\{10\}, then X/Y = [0,2) \cup (4,10) \cup \{t\}.

The topology on X/Y is just as you said:

U\subseteq X/Y is open iff p^{-1}(U) is open in X.

You are right that there is nothing to prove so far, because the fact that this defines a topology was probably proved in your book (it would be a very short proof anyway). In your problem, the instruction "use it to define a topology" simply means "re-state the definition of identification space/quotient space applied to this context."

The actual task that requires work is to prove for all subsets Z of X, p(Z) is open in X/Y if and only if Y\cup Z is open in X.

For example, in the above example, let Z=(3,5). Is p(Z) open? What if Z=(2,5)? Can you draw a "picture" of X/Y? I imagine an interval [0,10] made of string, then bunch up the portion from 2 to 4 into a single glob or knot that is one point, plus loop the endpoint at 10 over and glue it into the glob. So X/Y, if you move and turn it around, looks like an uppercase letter P.

 

[latentcorpse]

hi. i follow your "picture" of X/Y as a letter P orientated differently.

if Z=(3,5) then Y \cup Z = [2,5) \cup \{ 10 \}
what is the definition of "open" here? is it that open balls are contained within the set?

surely if we use Z=(2,5) then the intersection wtih Y will just be exactly the same as above?

also, for the sake of rigour, what would you write to actually define the topology on the identification space as it asks in the question?

 

[Billy Bob]

I would simply write, "Then define U\subseteq X/Y to be open iff p^{-1}(U) is open in X."


For the example, I meant (1,5) instead of (2,5).

Is Y \cup Z = [2,5) \cup \{ 10 \} open in [0,10]? No it isn't. The topology I'm using on [0,10] is the ordinary topology as a subspace of the real number line.

 

[latentcorpse]

ok so [2,5) union {10} isn't open because it's closed at 2

but if Z=(1,5)

then Y union Z = (1,5) union {10} and that would be open, yes?

how do i proceed from here?

 

[Billy Bob]

No, (1,5)\cup \{10\} is not open in the ordinary topology on [0,10].

So if Z=(1,5), then p(Z) is not open in X/Y. That is, (1,2) \cup (4,5) \cup \{t\} is not open in X/Y.

The examples are just to help with the intuition, and to test whether your proof steps look good.

Anyway, you have to prove for all subsets Z of X, p(Z) is open in X/Y if and only if Y\cup Z is open in X.

Part 1. Let Z be a subset of X. Assume p(Z) is open in X/Y. You must prove Y\cup Z is open in X.

Part 2. Let Z be a subset of X. Assume Y\cup Z is open in X. You must prove p(Z) is open in X/Y.

Try it. If you get stuck, turn to the definitions. Use the two examples Z=(1,5) and Z=(3,5) for intuitive insight.

 

[latentcorpse]

why is (1,5) union {10} not open? is it because the topological space is [0,10] and so {10}, despite being a singleton which are usually all open, in fact includes one of the closed endpoints of the space?

 

[latentcorpse]

if p(Z) is open in X/Y then by defn of a topology p^{-1}(p(Z)) is open in X. i think i know what i want to write next just not sure how to write it... we need to show p^{-1}(p(Z))=Y \cup Z since if Z \cap Y \neq \emptyset then p(Z)=Z \cup \{ t \} and then p^{-1}(p(Z))=Z \cup Y
however I'm struggling with the case of Z \cap Y = \emptyset. here p(Z)=Z and p^{-1}(p(Z))=Z again and so this tells us Z is open but doesn't guarantee that Y \cup Z is going to be open, does it?

 

[Billy Bob]

Yeah, I'm having the same problem as you. Re-read the problem as it was given to you. Are you missing something in the statement? Was Y supposed to be open? Was Z supposed to be assumed to contain at least one point of Y?

Regarding your earlier question about singleton sets, they are closed but not open in the ordinary topology (think of the definition of the usual topology using open intervals).

 

[latentcorpse]

i wrote out the problem word for word in my original post - there's no mention of Y being open or Z containing part of Y...

can you explain that singleton thing in more depth. do u just mean that the single point [5] on the real line would be closed and not open? are singletons always closed?

 

[Billy Bob]

On the real line, S={5} is closed because its complement is open.

S={5} is not open, because it is not the case that there exists an open interval I such that 5\in I\subseteq S.

I think the original problem is not correct as stated. Z=(5,6) in my earlier example is open in X, and p(5,6)=(5,6)=U is open in X/Y since p^{-1}(U)=(5,6) is open in X. However, Y\cup Z is not open in X.

 

[latentcorpse]

why is Z union Y not open in X in that case?

also, before i ask the lecturer, we agreed that this would work if it said in the question that Y was open. However, we are told that Y is a subset of the topological space X. By definition, a topology on a set X is a collection of of subsets that satisfy the three standard axioms. we can write X as a collection,U, of two subsets: Y and X\Y.
(i) both X, \emptyset \in U
(ii) the intersection Y \cap X \backslash Y = \emptyset and the empty set is open
(iii) their union is X itself and X is open
hopefully you'll verify that the above works which will guarantee Y is an open subset.

 

[latentcorpse]

bump.

 

[Billy Bob]

If Z=(5,6) in my example, then Y\cup Z=[2,4]\cup(5,6)\cup\{10\} which is definitely not open.

In your last question, you proposed defining a topology on X where the only open sets are X, the empty set, Y, and X-Y. That would be a topology, but it is irrelevant to this problem. The problem starts with a given topology on X, so you have to use the given arbitrary topology on X, not one of your own choosing.

 

[latentcorpse]

isn't it the case that if we can show Y obeys those three axioms then it will automatically be an open set and part of the topology?

 

[Billy Bob]

isn't it the case that if we can show Y obeys those three axioms then it will automatically be an open set and part of the topology?

No. The axioms in the definition of a topology describe the collection of sets that are open, not any individual open set. In your problem, a particular but arbitrarily chosen set X is given, along with a given particular but arbitrarily chosen collection of subsets (called open sets) that satisfy the axioms for being a topology. You can't change the collection of open sets to be something else.

I bet correct statement of the problem assumes Z\cap Y\neq\varnothing. There could be an alternate conclusion for the case Z\cap Y=\varnothing.

 

[latentcorpse]

ok. i'll email the lecturer about this then.

if you're telling me that we have X along with an arbitrary collection of subsets called the open sets. what is wrong with showing Y is one of these open sets by showing it satisfies the axioms for being a topology?

 

[latentcorpse]

yes we are to assume that Z \cap Y \neq \emptyset

this means i can just use the first half of my argument in post #12 to show Z \cup Y is open in X, yes?

that provides the \Rightarrow part of the iff statement

now consider the \Leftarrow part:

assume Z \cup Y is open in X. how do i prove p(Z) is open in X \backslash Y?
i know p(Z)=Z \cup \{ t \} but how do i prove this is open?

 

[latentcorpse]

bump.

 

[Billy Bob]

For both directions ("if" and "only if"), all you have to do now is show p^{-1}(p(Z))=Z\cup Y. You could try proving \subseteq and \supseteq if you think this statement requires proof.

But note that p(Z) is not Z\cup \{t\}.

 

[latentcorpse]

hold on.
in the direction where i assume p(Z) is open in X/Y. i know p^{-1}(p(Z)) is going to be open in X be definition of an identification space. so i say that because Z \cap Y \neq \emptyset, p(Z)=Z \cup \{ t \} \Rightarrow p^{-1}(p(Z))=Z \cup Y hence showing Z \cup Y is open in X.
but your saying i can't do this because p(Z) \neq Z \cup \{ t \}. Why not? will p(Z)=(X \cap Z) \cup \{ t \} instead? then p^{-1}(p(z))=(X \cap Z) \cup Y = (X \cup Y) \cap (Z \cup Y) = Z \cup Y. is this better?

then for the other direction, surely i need to get from the assumption that Z \cup Y is open in X to showing p(Z) is open in X/Y. i don't know how to do this?

 

[Billy Bob]

will p(Z)=(X \cap Z) \cup \{ t \} instead? then p^{-1}(p(z))=(X \cap Z) \cup Y = (X \cup Y) \cap (Z \cup Y) = Z \cup Y. is this better?

I guess I would call it "better" but it is still not right. Test your statements against the example of X=[0,10], Y=[2,4]\cup\{10\}, and Z=(3,5). I predict you can easily fix your statements. In both attempts, I believe you were thinking along the right lines, but your statements were not correct. The problem is that points like 3.5 (i.e. 3 1/2) are in Z but are not even in X/Y.

then for the other direction, surely i need to get from the assumption that Z \cup Y is open in X to showing p(Z) is open in X/Y. i don't know how to do this?

Once you show p^{-1}(p(Z))=Z\cup Y (which is true regardless of whether or not p(Z) is open, provided Z\cap Y\neq\varnothing) then simply use the definition of "open" in the identification space. There really is nothing left to prove at this point.

 

[latentcorpse]

how about

p(Z)=(X \backslash Y \cap Z) \cup \{ t \}
p^{-1}(p(Z))=(X \backslash Y \cup Y) \cap (Z \cup Y) = Z \cup Y?

also, is it true that your saying all i have to do is show p^{-1}(p(Z))=Z \cup Y and not worry about the if and only if statement because the defn of identification spaces when it talks about p(Z) being open if and only if p^{-1}(p(Z)) is open i.e. the if and only if statement is taken care of by the defn of the identification space?

 

[Billy Bob]

Looks good!

 

[latentcorpse]

so that's the whole question done yeah?