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I'm asked to prove the statement in the title under the assumption that I do not know the Inclusion-Exclusion Principle. I have two ways of starting the proof where:

  1. I could declare two sets with a certain amount of values and show by example that it is true:

$A = \{1. 2, 3, 4\}$ and $B = \{3, 4, 5, 6\}$

$|A| = 4$, $|B| = 4$

$|A \cup B| = |A| + |B| - |A \cap B| = 4 + 4 - 2 = 6$

  1. I could state that it is true and give a logical explanation:

This is true, because to count the number of elements in $A \cup B$, we start by counting those in $A$, and then add those in $B$. If $A$ and $B$ were disjoint, then we are done, otherwise, we have double counted those in both sets, so we must subtract those in $A \cap B$.

However, I don't know if these are counted as formal proofs. If not, how would I start a proof like this?

J.-E. Pin
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Random Student
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    "I could declare two sets with a certain amount of values and show by example that it is true" No. No. Certainly not. This is fine for building intuition only. Otherwise, all you accomplished in doing so would be showing that it happened to be true for that one specific example. Showing it is true for that one example however does not directly imply it is true for all of the infinitely many other examples out there. – JMoravitz Sep 21 '20 at 12:58
  • An example does not constitute a proof, so you need a suitable explanation. Perhaps you could start by considering sets A an B, and the number of elements for each of the possibilities A only, B only, and both A and B. – Peter Sep 21 '20 at 12:59
  • My unsophisticated notion is that given 2 sets, A and B, for each element there are 4 possibilities: either the element is or is not in A, either the element is or is not in B. However, I never had formal set theory training. For a formal approach, my guess is that Jean Marie's answer is the way to go. – user2661923 Sep 21 '20 at 13:03
  • To emphasize the "an example is not proof" there are several mathematical statements who happen to be true for the first many examples. Checking a large number of finite examples and they all happened to may lead you to believe the statement is true always. However, that may just be because you didn't check enough and a counterexample still exists. See conjectures that have been disproven with extremely large counterexamples. – JMoravitz Sep 21 '20 at 13:33
  • I see, so examples are not a formal proofing method. For future reference, would explaining it like in the 2nd option be a viable proofing method? – Random Student Sep 21 '20 at 13:36

1 Answers1

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Preliminary: I will use the rather classical notation: $E \backslash F$ for the set of elements that belong to $E$ but not to $F$.

You need to write $A \cup B$ as a disjoint union (said otherwise, a partition of $A \cup B$):

$$A \cup B=(A \backslash B) \cup (A \cap B) \cup (B \backslash A)$$

to which you can apply:

$$|A \cup B|=|A \backslash B| + |A \cap B| + |B \backslash A| \tag{1}$$

For the same reason, you can say that:

$$|A \cup B|=|A| + |B \backslash A| \tag{2}$$

$$|A \cup B|= |A \backslash B|+|B| \tag{3}$$

Now make the combination (2)+(3)-(1).

Jean Marie
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