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

Posted: Tue Dec 20, 2011 9:51 am
by yoyobarn
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 $$x\in A' \cup B'$$,
$$x\in cl(A-\{x\})\cup cl(B-\{x\})=(A-\{x\})\cup (A-\{x\})'\cup (B-\{x\})\cup (B-\{x\})'$$
$$\therefore x\in (A\cup B-\{x\})$$

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

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

Posted: Tue Dec 20, 2011 9:57 am
by yoyobarn
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?

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

Posted: Tue Dec 20, 2011 12:47 pm
by owlpride
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.

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

Posted: Tue Dec 20, 2011 5:54 pm
by fireandgladstone
I think A' = X\{pt}.

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

Posted: Tue Dec 20, 2011 6:04 pm
by owlpride
^ Thanks, you are right. That makes me feel a lot better.

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

Posted: Tue Dec 20, 2011 6:08 pm
by fireandgladstone
If A = {pt} then A\{x} = {pt} if x \neq pt. The derived set need not be closed.

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

Posted: Tue Dec 20, 2011 6:14 pm
by fireandgladstone
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).

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

Posted: Wed Dec 21, 2011 3:49 am
by yoyobarn
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.

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

Posted: Sun Aug 11, 2013 11:23 am
by Ellen
What is {pt} referencing? Pointless topology?

Thanks.

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

Posted: Wed Sep 18, 2013 3:38 pm
by goatman2743
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)?