Fastest Way to solve 9367 Q56 (without memorizing the ans)

Forum for the GRE subject test in mathematics.
yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby yoyobarn » Tue Dec 20, 2011 9:51 am

Hi, anyone has any practical tips to solve this type of topology questions within two minutes?

My long way of proving is as follows:

I) Let x \in (A\cup B)'.
x\in cl(A\cup B-\{x\}).
x\in (A\cup B-\{x\})\cup(A\cup B-\{x\})'
Case 1: If x\in A, then x\in (A-\{x\})\cup(A-\{x\})'=cl(A-\{x\})
\therefore x\in A'
Case 2: If x\in B, then similarly x\in B'.
\therefore x\in A' \cup B'
\therefore (A\cup B)'\subseteq A'\cup B'

Conversely, if x\in A' \cup B',
x\in cl(A-\{x\})\cup cl(B-\{x\})=(A-\{x\})\cup (A-\{x\})'\cup (B-\{x\})\cup (B-\{x\})'
\therefore x\in (A\cup B-\{x\})

\therefore x\in (A\cup B-\{x\})\cup (A\cup B-\{x\})'=cl(A\cup B-\{x\})
\therefore x\in (A\cup B)'
\therefore A'\cup B'\subseteq (A\cup B)'

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby yoyobarn » Tue Dec 20, 2011 9:57 am

II)
Let A=[0,1).
B=(1,2].

Then LHS=(\emptyset)'=\emptyset

RHS=[0,1]\cap [1,2]=\{1\}.

III)
cl(A)=A\cup A'=A\cup\emptyset=A
Therefore, A is closed in X.

IV) Let A=\emptyset, which is open.
A'=\emptyset

Is the above proof acceptable?
Also, any tips for solving this kind of questions within the time limit in the GRE. Any tricks like drawing circles, diagrams, powerful theorems that solve it instantly?

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby owlpride » Tue Dec 20, 2011 12:47 pm

I believe that III is false. Suppose that X has the trivial topology and A = {pt}. Then A' is empty but A is not closed in X. The formula cl(A) = A union A' only holds for sufficiently nice spaces (e.g. Hausdorff spaces).

I believe that only I is correct, which is problematic because it's not an answer choice.

fireandgladstone
Posts: 27
Joined: Sun Oct 17, 2010 4:57 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby fireandgladstone » Tue Dec 20, 2011 5:54 pm

I think A' = X\{pt}.

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby owlpride » Tue Dec 20, 2011 6:04 pm

^ Thanks, you are right. That makes me feel a lot better.
Last edited by owlpride on Tue Dec 20, 2011 6:08 pm, edited 1 time in total.

fireandgladstone
Posts: 27
Joined: Sun Oct 17, 2010 4:57 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby fireandgladstone » Tue Dec 20, 2011 6:08 pm

If A = {pt} then A\{x} = {pt} if x \neq pt. The derived set need not be closed.

fireandgladstone
Posts: 27
Joined: Sun Oct 17, 2010 4:57 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby fireandgladstone » Tue Dec 20, 2011 6:14 pm

Also, if you're going for speed, notice that you only needed to check claim 2 to get the answer (based on the available answers).

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby yoyobarn » Wed Dec 21, 2011 3:49 am

fireandgladstone wrote:Also, if you're going for speed, notice that you only needed to check claim 2 to get the answer (based on the available answers).


Wow! This is a really useful tip! Thanks.

Ellen
Posts: 3
Joined: Sat Aug 10, 2013 3:56 pm

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby Ellen » Sun Aug 11, 2013 11:23 am

What is {pt} referencing? Pointless topology?

Thanks.

goatman2743
Posts: 31
Joined: Mon Sep 02, 2013 7:29 pm

Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)

Postby goatman2743 » Wed Sep 18, 2013 3:38 pm

yeah, the only way I could think to solve this was by eliminating II) and noticing from the choices that the answer has to be b. However, going back and looking at it I still don't see how I) can be eliminated, I think there is a problem with op's proof that (A \cup B)' \subseteq A' \cup B'. op splits it up into the cases x \in A and x \in B but we can't conclude x \in A \cup B from x \in (A \cup B)'. For example, under the indiscrete topology, (A \cup B)' is the entire set.

Am I right here, and if so does anyone know of a proper proof for I)?




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