If A and B are a subset of C. I know it’s true by showing examples but how is it proved?

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Apply the definitions.

– arkeet

Oct 21 at 0:41

@arkeet you mean the definition of union? It’s the fact that they are inside a function which is throwing me off. Because if you take it out of the function it’s just A U B = A U B?

– James Mitchell

Oct 21 at 0:43

1

(Let f:X→Yf \colon X \to Y.) To show e.g. that f(A∪B)⊆f(A)∪f(B)f(A \cup B) \subseteq f(A) \cup f(B), assume y∈f(A∪B)y \in f(A \cup B) and show it’s in the latter set. If you look at the definition of f(A∪B)f(A \cup B), you will see it is the set of all y∈Yy \in Y such that y=f(x)y = f(x) for some x∈A∪Bx \in A \cup B. That is the start.

– arkeet

Oct 21 at 0:45

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2 Answers

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Prove both inclusions.

f(A∪B)⊆f(A)∪f(B):f(A \cup B) \subseteq f(A) \cup f(B):

Let xx in f(A∪B)f(A \cup B), there there exists y∈A∪By \in A \cup B such that f(y)=xf(y) = x. If y∈Ay \in A, then x∈f(A)x \in f(A), and if y∈By \in B, then x∈f(B)x \in f(B), so in both cases x∈f(A)∪f(B)x \in f(A) \cup f(B).

f(A)∪f(B)⊆f(A∪B):f(A) \cup f(B) \subseteq f(A \cup B):

Let x∈f(A)∪f(B)x \in f(A) \cup f(B).

If x∈f(A)x \in f(A), then there exists y∈Ay \in A such that f(y)=xf(y) = x. As yy is in AA, it is in A∪BA \cup B, so f(y)=x∈f(A∪B)f(y)=x \in f(A \cup B). If x∈f(B)x \in f(B), the procedure is analogous.

Hints:

First prove ⊃\supset. For that, observe A,B⊂A∪BA, B\subset A\cup B.

Then prove ⊂\subset taking an element y=f(x)y=f(x), x∈A∪B\;x\in A\cup B. Where can yy dwell?

What is the backwards subset sign?

– James Mitchell

Oct 21 at 0:49

Contains (as a subset, not as an element).

– Bernard

Oct 21 at 1:00