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.

Many thanks for your help.