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Find a counterexample to show that the following statement is false. For all sets A and B, (A ∪ B)c = Ac ∪ Bc. Assume that all sets are subsets of a universal set U = {1, 2, 3, 4, 5}. (Enter your answer as a comma-separated list in the form A, B where both A and B are written in set-roster notation. Enter EMPTY or ∅ for the empty set.)

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Answer:

Given,

The universal set,

U = {1, 2, 3, 4, 5},

Let A = {1, 2},

B = {2, 3, 4}

[tex]A\cup B=\text{ all elements of A and B }=\{1, 2, 3, 4\}[/tex]

[tex](A\cup B)^c=U-(A\cup B) = \{1, 2, 3, 4, 5\} - \{1, 2, 3, 4\} = \{5\}[/tex]

Now,

[tex]A^c = U - A = \{3, 4, 5\}[/tex]

[tex]B^c=U-B=\{1, 5\}[/tex]

[tex]A^c\cup B^c = \{1, 3, 4, 5\}[/tex]

[tex]\implies (A\cup B)^c\neq A^c\cup B^c [/tex]

Hence, proved......

facundo3141592

The counterexample is:

A = {1} and B = {1, 2}

the statement is true only if the intersection of the sets is empty.

How to find a counterexample?

We can define:

A = { 1 }

B = {1 , 2}

The union of A and B is:

A U B = {1, 2}

So we have:

A U B = B

Now, the elements that do not belong to the union are:

(A U B)c = U - (A U B)

Where U = {1, 2, 3, 4, 5}

Then we have:

(A U B)c = {1, 2, 3, 4, 5} - {1, 2} = {3, 4, 5}

If instead we take the union of Ac and Bc we get:

  • Ac = U - A = {1, 2, 3, 4, 5} - {1} = {2, 3, 4, 5}
  • Bc = U - B = {1, 2, 3, 4, 5} - {1, 2} = {3, 4, 5}

Then:

Ac U Bc =  {2, 3, 4, 5} ≠  {3, 4, 5} = (A U B)c

So we just found two sets A and B such that the statement is false, this is the counterexample.

If you want to learn more about counterexamples, you can read:

https://brainly.com/question/13458417

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