This would be a false statement, correct?

  • #1
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Summary:

A quick question about sets and intersection/unions

Main Question or Discussion Point

I believe that I am correct, the following statement here must be FALSE, right? It has to be false because A union B is like the two entire circles of the Venn diagram and that cannot be a subset of the intersection area, right? Now if this statement was flipped, then it would be true?

31826703-6ACD-452A-94E8-02CDF6802FFF.jpeg
 

Answers and Replies

  • #2
BvU
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Correct :wink: !
 
  • #3
BvU
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By the way PF would really appreciate typed posts instead of disk-space wasting big pictures !

If you type
##A\cup B \subseteq A\cap B## then you get ##A\cup B \subseteq A\cap B## (in-line math), and if you type
$$A\cup B \subseteq A\cap B$$ then you get $$A\cup B \subseteq A\cap B$$
('displayed math').

##LaTeX## is fun, fairly easy to start with and extremely useful
 
  • #4
Math_QED
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Yes correct. For example take ##A = \{0\}, B = \{1\}##. Then ##A\cup B = \{0,1\}## yet ##A \cap B = \emptyset##.

The other inclusion, that is ##A \cap B \subseteq A \cup B##, is trivially true.
 
  • #5
PeroK
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Summary:: A quick question about sets and intersection/unions

I believe that I am correct, the following statement here must be FALSE, right? It has to be false because A union B is like the two entire circles of the Venn diagram and that cannot be a subset of the intersection area, right? Now if this statement was flipped, then it would be true?
You can, however, have ##A \cup B = A \cap B##, so it's not always false. Can you see when you have this equality?

It is correct that ##A \cup B## can never be a proper subset of ##A \cap B##.
 
  • #6
pbuk
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You can, however, have ##A \cup B = A \cap B##, so it's not always false. Can you see when you have this equality?
Yes, absolutely, it is NOT correct to say that the statement is unconditionally false.
 
  • #7
Stephen Tashi
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I believe that I am correct, the following statement here must be FALSE, right?
To reinforce what others have said, there is a difference between talking about a "statement" versus taking about a "propositional function".

If you utter a phrase like "If x > 7 then x > 1" , it is technically a "propositional function", which cannot be assigned a truth value ( True vs False) until "x" is defined to be something specific. In common speech and in informal mathematical writing, it is usually understood that phrases like "If x >7 then x > 1" are intended to have the "universal quantifier" given by the phrase "for each". So the reader interprets the claim "If x > 7 then x > 1" to mean "For each number x, if x >7 then x > 1". With this interpretation, the phrase becomes a statement which can be assigned a truth value.

The other commonly used quantifier is "there exists". Someone writing hasty notes might jot down the phrase "x > 1" intending it to mean "There exists a number x such that x > 1". However, this is not a clear style of writing.

Your question about "##A \cup B \subseteq A \cap B##" is technically a question about a propositional function, so it cannot be assigned a single truth value. You probably intended some universal quantifiers to be supplied, so the phrase would become the statement "For each set ##A## and for each set ##B##, ## A \cup B \subseteq A \cap B##". With the universal quantifiers, the propositional function is converted into a False statement. However, if we supply existential quantifiers, we get "There exists a set ##A## and there exists a set ##B## such that ## A \cup B \subseteq A \cap B##". This converts the propositional function into a True statement.
 

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