## Topology reflections

Forum for the GRE subject test in mathematics.
lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Topology reflections

Although during the test there are just probably a couple or triple of questions concerning topology, for me it is the most enigmatic and interesting topic among others. Nevertheless, there are some questions that are still baffling me.

1. Let's consider the set of real numbers R and two topologies defined on it:
T1 - standard real line topology;
T2 - lower-limit topology.
Therefore, I see that:
T1 = {empty set, R, all intervals (a,b)};
T2 = {empty set, R, all intervals [a,b)}.
At the same time, it is considered that T2 is strictly finer than T1. That actually implies that T1<T2 which assumes that T2 contains all elements of T1. But T2 does not have intervals (a,b)? Does it?

2. Let's consider again the set R of real numbers and the lower-limit topology T2. Apparently, with this topology R is Hausdorff space, since every two distinct points can be separated by neighborhoods. There is a property that
Every singleton in Hausdorff space is closed.
At the same time, according to the definition of closed set as complement of open set, I can see that only closed set in T2 are just
empty set and R (they are actually both open and closed in every topology);
[a,b) (they are actually both open and closed in T2);
(-inf,a).
But I can't think of example where singleton could be a complement of open set in T2. Unless, as it was discussed in question 1, the intervals (a,b) are also open in T2.

What you think about that? lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Dear friends, suddenly I have discovered for myself that intervals (a,b) are really open in the lower-limit topology. Indeed, each such interval is the union of infinitely number of intervals [a+1/n, b) for n=[k,+inf]. Therefore, both my questions are answered. Nevertheless, this fact just produces my another misconception in answering the next question:

3. Let R be the set of real numbers with the topology generated by the basis {[a,b): a<b, where a,b - real numbers}. If X is the subset [0,1] of R, which of the following must be true?

I. X is compact.
II. X is Hausdorff.
III. X is connected.

A. I only
B. II only
C. III only
D. I and II
E. II and III

Is is obvious that this set is Hausdorff and connected. Now I can see also that that since intervals (-inf, a) and (b;+inf) are open in described topology, X must be closed! Therefore, using the fact that it is also bounded I conclude that it is compact as well.
Nevertheless, the right answer is B (!!) which implies that it is not compact!
My first guess was that "closed and bounded" properties are not working for low-limit topology. But this interval still looks for me like it's every open covering will always contain finite subcollection.

Who is wrong - the answer or I?

DesolateReality
Posts: 1
Joined: Fri Mar 14, 2008 2:29 am

### back to definitions!

the lower limit topology on the reals is not compact. The following is an open covering with no finite subcover:
{[n,n+1): n is an integer}.

the thing is that only in the standard topology of the reals, does the heine-borel theorem holds. That is, only for the reals with the standard topology, which you call T1, are the following two facts about a subset A of R equivalent:
1) A is compact,
2) A is closed and bounded.
Since you are working with T2 for the multiple-choice question, this theorem does not hold anymore.

Also, the lower limit topology on the reals result in a disconnected topology. This is because both [0,infinity) is closed and open.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
the lower limit topology on the reals is not compact. The following is an open covering with no finite subcover:
{[n,n+1): n is an integer}.
Cool! And here is the covering of [0,1] interval that has no finite subcover:
{[0,1-1/n), n is an integer} + [1,2)

Now I also see that Heine-Borel theorem works only to standard topology. Thanks for answer. Continuing the topology discussion I would come up with the next question.

4. Is the set of points in the 2D space such that x in [a,b], y=0 compact or not? Why?

JcraigMSU
Posts: 13
Joined: Wed Aug 06, 2008 9:25 am
I was wondering how [n, n+1) is an open cover. i am new to topology, haha. If you were at x = n and subtracted epsilon, you would be outside of the cover?

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
I was wondering how [n, n+1) is an open cover?
It is an open cover in the lower-limit topology.

mathsubboy
Posts: 11
Joined: Sun Jul 20, 2008 12:11 pm
As for lime's problem 4:

under the standard topology, any set is compact if and only if it is both closed and bounded, whatever the dimension is.

so the set of points in the 2D space such that x in [a,b], y=0 is compact.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
Is the set of points in the 2D space such that x in [a,b] OR y=0 compact or not? Why?

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Is the set of points in the 2D space such that x in [a,b] OR y=0 compact or not? Why?
If you mean union of X axis and set {(x,y): x in [a,b] y in R} then it is not compact as the union of two unbounded sets.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
Now, let discuss about topology :

Problem #62 GR0568
Let K be a nonempty subset of R^n. Which of the following is true :

1. If K is compact, then every real-continuous function on K in bounded
2. If every real-continuous function on K is bounded then K in compact
3. If K is compact then K is connected

1 & 3 are cleared, I have problem with 2

If some one knows , please explain? thanks

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
Hi dear,

No one want to discuss with me mathsubboy
Posts: 11
Joined: Sun Jul 20, 2008 12:11 pm
2 is correct,
firstly, K must be bounded, otherwise you can construct a unbounded continuous function on it
Secondly, K must not be open , or again you can find a counterexample.

So K must be bounded and closed, that is compact.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
for (2), K is bounded is obvious ( let f: K------->K with f(x)=x, them K is bounded)

Now prove that K is closed. Suppose that K is not closed, then R^n\K is not open : since K is bounded so K must have form :
K=(a1,b1)x(a2,b2]x....x(an,bn)
so let F : K--------> R F(x1,x2,....,xn)=[1/(x1-a)]x2....xn then f is continuous and f is not bounded - contradiction

so K <R^n is closed and bounded then K is compact