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kotegsom [21]
3 years ago
8

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

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

(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.

<h3>(a)</h3>

Note that A \cap B can well be smaller than A. 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.

<h3>(b)</h3>

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.

<h3>(c)</h3>

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

Let x 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 x 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, OR
  • both 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).

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