# The word "or" in set theory

1. Jan 6, 2016

### A.MHF

1. The problem statement, all variables and given/known data
So I was doing this problem in Munkres's Topology book:

Determine whether the statement is true or false, If a double implication fails, determine whether one or the other of the possible implications holds:

A ⊂ B or A ⊂ C ⇔ A ⊂ ( B ∪ C )

2. Relevant equations
-

3. The attempt at a solution
I know that the ⇒ direction is true because (x∈A→x∈B or x∈A→x∈C)⇒(x∈A→(x∈B or x∈C))
For the other direction, I thought at first that it's true, but I checked some online answers and what I found is it's false. I thought it's true because what I had in mind is the word "or" stands for "either A is a subset of B, or A is a subset of C, or both". And if A is a subset of the union of B and C, then it's implied that A is a subset of at least one of them. What am I getting wrong?

This is the answer that I found online:

If the LHS statement is true then for each we have that and so , i.e. . The other direction is not correct though. Consider the sets then but it is not true that .

2. Jan 6, 2016

### Samy_A

The example shows that this is wrong: "And if A is a subset of the union of B and C, then it's implied that A is a subset of at least one of them."
A={1,3}⊂ B∪C={1,2,3,4}, but clearly A is not a subset of either B or C.

3. Jan 6, 2016

### Staff: Mentor

Your problem isn't in misunderstanding of the word "or" (as implied by your thread title). The counterexample given below clearly shows that A can be contained in $B \cup C$ without A being a subset of either of the other sets.

Note that in the example, one element of A is contained in B, and the other element of A is contained in C, but A is not a subset of either B or C.

4. Jan 6, 2016

### A.MHF

Oh, I don't how I missed that, feel stupid for asking haha.
Thanks.