Forum for the GRE subject test in mathematics.
chival
Posts: 6
Joined: Wed Nov 04, 2009 1:31 am

Hey everyone,

I don't know if I have found a paradox, look at this problem (GR9367 #56):

2. For a subset S of a topological space X, let cl(S) denote the closure of S in X, and let S'={x:x in cl(S-{x})} denote the derived set of S. If A and B are subsets of X, which of the following statements are true?
I. (AuB)'=A'uB'
II. (AnB)'=A'nB'
III. If A' is empty, then A is closed in X.
IV. If A is open in X, then A' is not empty.

A. I and II only
B. I and III only
C. II and IV only
D. 1, II, and III only
E. I, II, III, and IV

The answer is B, and this is the explanation posted by CoCoA:

item I: true: if x in (AuB)', then x in cl(AuB-{x}), so x in cl(A-{x}) u cl(B-{x}) because cl(AuB-{x}) is the smallest closed set containing AuB-{x}, thus x in A'uB'. if x in A'uB', then assume that x in A', so x in cl(A-{x}), thus x in cl(AuB-{x}), therefore x in (AuB)'.

item III: true: because then cl(A)=AuA'=A so A is closed.

item IV: false: if X has the discrete topology then there are NO limit points and every subset is open.

My question is, in the sense of relation, IV seems to be contrapositive to III (R->P == !p->!R):

If A is empty, then B is closed.
If B is open, then A is not empty.

So I don't understand why III could be true whilest IV is false.

chival
Posts: 6
Joined: Wed Nov 04, 2009 1:31 am

Topology can be so difficult to understand sometimes.. i got another one:

Does f(x) being a continous map make f-1(x) the inverse function an open map?

after reading the definitions i cant help wondering if these two are in fact the same thing

feroz_apple
Posts: 30
Joined: Sat Oct 17, 2009 1:20 pm

chival wrote:Topology can be so difficult to understand sometimes.. i got another one:

Does f(x) being a continous map make f-1(x) the inverse function an open map?

after reading the definitions i cant help wondering if these two are in fact the same thing

true...but only if the inverse function exists and i hope u know that not every cont. func. is bijective
Last edited by feroz_apple on Wed Nov 04, 2009 10:52 am, edited 1 time in total.

feroz_apple
Posts: 30
Joined: Sat Oct 17, 2009 1:20 pm

chival wrote:Hey everyone,

I don't know if I have found a paradox, look at this problem (GR9367 #56):

2. For a subset S of a topological space X, let cl(S) denote the closure of S in X, and let S'={x:x in cl(S-{x})} denote the derived set of S. If A and B are subsets of X, which of the following statements are true?
I. (AuB)'=A'uB'
II. (AnB)'=A'nB'
III. If A' is empty, then A is closed in X.
IV. If A is open in X, then A' is not empty.

A. I and II only
B. I and III only
C. II and IV only
D. 1, II, and III only
E. I, II, III, and IV

The answer is B, and this is the explanation posted by CoCoA:

item I: true: if x in (AuB)', then x in cl(AuB-{x}), so x in cl(A-{x}) u cl(B-{x}) because cl(AuB-{x}) is the smallest closed set containing AuB-{x}, thus x in A'uB'. if x in A'uB', then assume that x in A', so x in cl(A-{x}), thus x in cl(AuB-{x}), therefore x in (AuB)'.

item III: true: because then cl(A)=AuA'=A so A is closed.

item IV: false: if X has the discrete topology then there are NO limit points and every subset is open.

My question is, in the sense of relation, IV seems to be contrapositive to III (R->P == !p->!R):

If A is empty, then B is closed.
If B is open, then A is not empty.

So I don't understand why III could be true whilest IV is false.

3) is true because if A' is empty then as cl(a) = A union A' = A. So A is closed
Now 4) is false because
consider the space (X,T) where X = {x,y} and topology T = {phi,X,{x}}
Let A = {x}
then i believe you can see that no such element "a" exists such that a is in cl(A-{a})
or A' = phi.
Hence A is empty but A' is phi

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

I think the confusion stems from "not open" not meaning "closed", so those aren't contrapositive of each other.

chival
Posts: 6
Joined: Wed Nov 04, 2009 1:31 am

feroz_apple wrote:
chival wrote:Topology can be so difficult to understand sometimes.. i got another one:

Does f(x) being a continous map make f-1(x) the inverse function an open map?

after reading the definitions i cant help wondering if these two are in fact the same thing

true...but only if the inverse function exists and i hope u know that not every cont. func. is bijective

so here is the tricky thing, why would it be defined in the first place, if the inverse function f-1(x) is not guaranteed to exist, that the map would a continuous map only on the condition that the f-1(x) is open in X1 for every open subset O in X2?

and i think the logic is a lil fuzzy here, because according to another theory, if the inverse function f-1(x) does exist, it is guaranteed to be continuous, which also ensured the openness of both f(x) and f-1(x)

i don't know if i'm losing it now... my test is scheduled tomorrow

feroz_apple
Posts: 30
Joined: Sat Oct 17, 2009 1:20 pm

feroz_apple wrote:
chival wrote:Topology can be so difficult to understand sometimes.. i got another one:

Does f(x) being a continous map make f-1(x) the inverse function an open map?

after reading the definitions i cant help wondering if these two are in fact the same thing

true...but only if the inverse function exists and i hope u know that not every cont. func. is bijective

so here is the tricky thing, why would it be defined in the first place, if the inverse function f-1(x) is not guaranteed to exist, that the map would a continuous map only on the condition that the f-1(x) is open in X1 for every open subset O in X2?

and i think the logic is a lil fuzzy here, because according to another theory, if the inverse function f-1(x) does exist, it is guaranteed to be continuous, which also ensured the openness of both f(x) and f-1(x)

i don't know if i'm losing it now... my test is scheduled tomorrow

Seriously mate ...ure losing it