Mathematics, 13.10.2020 23:01 vanleervanleer

Which of the following statements is true for all sets A, B and C? Give a proof or a counter example.

(a) A ⊆ ((A∩B)∪C).

(b) (A∪B)∩C = (A∩B)∪C.

(c) (A\B)∩C = (A∩C) \ (B∩C).

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Answer from: flamingtacos4026

(a) and (b) are not true in general. Refer to the explanations below for counterexamples.

It can be shown that (c) is indeed true.

Step-by-step explanation:

This explanation will use a lot of empty sets $\phi$ just to keep the counterexamples simple.

(a)

Note that $A \cap B$ can well be smaller than . It should be alarming that the question is claiming $A\!$ to be a subset of something that can be smaller than $\! A$ . Here's a counterexample that dramatize this observation:

Consider:

$A = \left\lbrace 1 \right\rbrace$ . $B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .) $C = \phi$ (another empty set.)

The intersection of an empty set with another set should still be an empty set:

$A \cap B = \left\lbrace 1\right\rbrace \cap \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

The union of two empty sets should also be an empty set:

$((A \cap B) \cup C) = \left\lbrace\right\rbrace \cup \left\lbrace\right\rbrace = \left\lbrace\right\rbrace$ .

Apparently, the one-element set $A = \left\lbrace 1 \right\rbrace$ isn't a subset of an empty set. $A \not \subseteq ((A\cap B) \cup C)$ . Contradiction.

(b)

Consider the same counterexample

$A = \left\lbrace 1 \right\rbrace$ . $B = \phi$ (an empty set, same as $\left\lbrace \right\rbrace$ .) $C = \left\lbrace 2 \right\rbrace$ (another empty set.)

Left-hand side:

$(A \cup B) \cap C = \left(\left\lbrace 1 \right\rbrace \cup \left\lbrace \right\rbrace\right) \cap \left\lbrace 2 \right\rbrace\right = \left\lbrace 1 \right\rbrace \cap \left\lbrace 2 \right\rbrace = \left\lbrace \right\rbrace$ .

Right-hand side:

$(A \cap B) \cup C = \left(\left\lbrace 1 \right\rbrace \cap \left\lbrace \right\rbrace\right) \cup \left\lbrace 2 \right\rbrace\right = \left\lbrace \right\rbrace \cup \left\lbrace 2 \right\rbrace = \left\lbrace 2 \right\rbrace$ .

Apparently, the empty set on the left-hand side $\left\lbrace \right\rbrace$ is not the same as the $\left\lbrace 2 \right\rbrace$ on the right-hand side. Contradiction.

(c)

Part one: show that left-hand side is a subset of the right-hand side.

Let be a member of the set on the left-hand side.

$x \in (A \backslash B) \cap C$ .

$\implies x\in A \backslash B$ and $x \in C$ (the right arrow here reads "implies".)

$\implies x \in A$ and $x \not\in B$ and $x \in C$ .

$\implies (x \in A\cap C)$ and $x \not\in B \cap C$ .

$\implies x \in (A \cap C) \backslash (B \cap C)$ .

Note that $x \in (A \backslash B) \cap C$ (set on the left-hand side) implies that $x \in (A \cap C) \backslash (B \cap C)$ (set on the right-hand side.)

Therefore:

$(A \backslash B) \cap C \subseteq (A \cap C) \backslash (B \cap C)$ .

Part two: show that the right-hand side is a subset of the left-hand side. This part is slightly more involved than the first part.

Let be a member of the set on the right-hand side.

$x \in (A \cap C) \backslash (B \cap C)$ .

$\implies x \in A \cap C$ and $x \not\in B \cap C$ .

Note that $x \not\in B \cap C$ is equivalent to:

$x \not \in B$ , OR $x \not\in C$ , ORboth $x \not\in B$ AND $x \not \in C$ .

However, $x \in A \cap C$ implies that $x \in A$ AND $x \in C$ .

The fact that $x \in C$ means that the only possibility that $x \not\in B \cap C$ is $x \not \in B$ .

To reiterate: if $x \not \in C$ , then the assumption that $x \in A \cap C$ would not be true any more. Therefore, the only possibility is that $x \not \in B$ .

Therefore, $x \in (A \backslash B)\cap C$ .

In other words, $x \in (A \cap C) \backslash (B \cap C) \implies x \in (A \backslash B)\cap C$ .

$(A \cap C) \backslash (B \cap C) \subseteq (A \backslash B)\cap C$ .

Combine these two parts to obtain: $(A \backslash B) \cap C = (A \cap C) \backslash (B \cap C)$ .

Answer from: Quest

if this is really you s much. i only have 3 points left and i have 2 questions to ask

Answer from: Quest

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

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