Fastest Way to solve 9367 Q56 (without memorizing the ans)
Fastest Way to solve 9367 Q56 (without memorizing the ans)
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 $$x\in A' \cup B'$$,
$$x\in cl(A\{x\})\cup cl(B\{x\})=(A\{x\})\cup (A\{x\})'\cup (B\{x\})\cup (B\{x\})'$$
$$\therefore x\in (A\cup B\{x\})$$
$$\therefore x\in (A\cup B\{x\})\cup (A\cup B\{x\})'=cl(A\cup B\{x\})$$
$$\therefore x\in (A\cup B)'$$
$$\therefore A'\cup B'\subseteq (A\cup B)'$$
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)
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?
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)
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.
I believe that only I is correct, which is problematic because it's not an answer choice.

 Posts: 27
 Joined: Sun Oct 17, 2010 4:57 am
Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)
I think A' = X\{pt}.
Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)
^ 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.

 Posts: 27
 Joined: Sun Oct 17, 2010 4:57 am
Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)
If A = {pt} then A\{x} = {pt} if x \neq pt. The derived set need not be closed.

 Posts: 27
 Joined: Sun Oct 17, 2010 4:57 am
Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)
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)
Wow! This is a really useful tip! Thanks.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).
Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)
What is {pt} referencing? Pointless topology?
Thanks.
Thanks.

 Posts: 31
 Joined: Mon Sep 02, 2013 7:29 pm
Re: Fastest Way to solve 9367 Q56 (without memorizing the ans)
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)?
Am I right here, and if so does anyone know of a proper proof for I)?