Understanding the or in Set Theory

A.MHF
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Homework Statement


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 )

Homework Equations


-

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
latex.png
we have that
latex.png
and so
latex.png
, i.e.
latex.png
. The other direction is not correct though. Consider the sets
latex.png
then
latex.png
but it is not true that
latex.png
.
 
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A.MHF said:

Homework Statement


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 )

Homework Equations


-

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
latex.png
we have that
latex.png
and so
latex.png
, i.e.
latex.png
. The other direction is not correct though. Consider the sets
latex.png
then
latex.png
but it is not true that
latex.png
.
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.

venn.jpg
 
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A.MHF said:

Homework Statement


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 )

Homework Equations


-

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?
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.
A.MHF said:
This is the answer that I found online:

If the LHS statement is true then for each
latex.png
we have that
latex.png
and so
latex.png
, i.e.
latex.png
. The other direction is not correct though. Consider the sets
latex.png
then
latex.png
but it is not true that
latex.png
.
 
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Likes A.MHF
Oh, I don't how I missed that, feel stupid for asking haha.
Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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