# Munkres HW Problem Help Regarding Quotient Spaces and Collapsing

1. Feb 25, 2014

### SetTheory

This is how the problem appears in my book(Munkres 2nd edition Topology, sect. 22 pg 145)

6. Recall that R_K denotes the real line in the K-topology. Let Y be the quotient space obtained from R_K by collapsing the set K to a point; let p : R_K → Y be the quotient map.

(a) Show that Y satisfies the T_1 axiom, but is not Hausdorff.

(b)Show that p x p: R_K x R_K → Y x Y is not a quotient map. [Hint: the diagonal is not closed in Y x Y, but its inverse image is closed in R_K x R_K]

And R_K is the topology with basis elements (a,b) and (a,b) - K, where K is the sequence 1/n|n is a natural number.

For (a) I am trying to prove that since the quotient space consists of a bunch of one-point sets and they are all closed, Y is T1. What does Y look like? I thought it was R_K where rather than subtracting the points in K, it was subtracting the collapsed single point set and then THIS set was being partitioned into the quotient space. Is this thinking correct, or does the subtraction of K come after R_K has been partitioned?

Assuming Y does look like this, then why must all these one point sets be closed? I remember that K is closed in R_K, but I do not see how this would imply all the one point sets to be closed. I think knowing how Y looks would help a lot.

For (b)...well I have an idea of how to approach this part.

2. Feb 26, 2014

### exclamationmarkX10

Y looks like this: there is a point, call it a, such that p(K) = {a}. for every other point x in R_K, p(x) = x.

here's a hint for part (a): if p is a quotient map, A is closed in Y iff the inverse image of A is closed in R_K (see the paragraph under the definition of quotient map in Munkres' book). to show it is not Hausdorff, consider 0 and K.

3. Feb 26, 2014

### SetTheory

Then does this mean that p is mapping all other points in R_K to themselves in Y? I suppose that must follow from the def. of quotient map. I guess I am confused about the purpose of collapsing K, if this hadn't been done would Y be Hausdorff?

4. Feb 26, 2014

### Dick

Yes, R_K is both T1 and Hausdorff. You might want to show this before you continue. The quotient space is still T1 but not Hausdorff. Consider the points 0 and $a$, as !!!!!!!!!! defined it. Can you show 0 has a neighborhood that does not contain $a$ and $a$ has a neighborhood that does not contain 0 but no such neighborhoods exist that don't intersect?

5. Feb 26, 2014

### SetTheory

Okay, yes it definitely makes sense why Y is not Hausdorff, because you cannot separate {a} from 0. One more question, I promise. It is to my understanding I can prove Y is T_1 by proving that each element of Y, the partition, is closed in R_K. I am inexperienced in proofwork, could you give me some direction on how to go about proving this?

Thanks for the help,

6. Feb 26, 2014

### exclamationmarkX10

You have to show: Every {y} where y is in Y is closed.

Hint: If p is a quotient map from X to Y, then A is closed in Y iff the preimage of A under p is closed.

7. Feb 26, 2014

### Dick

The only problematic points are $a$ and 0. Can you give an example of a neighborhood of 0 than does not contain $a$ (that should be easy) and a neighborhood of $a$ that does not contain 0 (that might be just a little harder).

8. Feb 26, 2014

### SetTheory

Okay, thank you! I am excited to fill out this proof now