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The question says :

If $f:X\rightarrow Y$ and $a,b \subseteq X$, then $f(a \cap b)$ is equal to?

- $f(a)-f(b)$
- $f(a)\cap f(b)$
- a proper subset of $f(a)\cap f(b)$
- $f(b)-f(a)$

I'm having a hard time understanding the question. I'm not sure what $a,b$ are because of the lowercase. Are they sets? And if they are, what is the question asking? Can someone translate the function in English?

Asaf Karagila
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Siddharth Thevaril
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  • Yes, $a$ and $b$ are sets. In general, $f(c) = {y \in Y : y = f(x),\ x \in c}$. – Umberto P. Feb 04 '15 at 17:14
  • Refer to 23. here. What a horrible notation. I swear I didn't understand what was meant before Umberto spelled it out. I was already thinking in transitive sets where $\subseteq$ can be interpreted as $\in$ and stuff like that to make sense of the question. – Git Gud Feb 04 '15 at 17:19

4 Answers4

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First, notice that $a,b$ are subsets of $X$, there. I would be more comfortable using $A$ and $B$ instead, so it is always clear that we're dealing with sets. You're supposed to use the notion of direct image: if $f: X \to Y$ is a function, and $A \subset X$, then: $$f(A) := \{ f(x) \in Y\mid x \in A \}.$$

You must check which of the set identities are valid or not.

Ivo Terek
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Yes, $a$ and $b$ are sets; they are subsets of $X$. The expression “$f(a)$” is an abuse of notation; in this question it means the set of values that are obtained by applying $f$ to some element of the set $a$. That is, $f(a)$ is being used as a shorthand for $\{ f(x) \mid x\in a\}$, which in turn is a standard shorthand for $\{ y \mid \exists x\in a. y=f(x)\}$.

The function $f$ is not given. However, there is a general property that holds for any function $f$, which relates $f(a\cap b)$ to $f(a)$ and $f(b)$. Your task is to discover this general property which is true of every function $f$.

Suppose someone told you to prove, or disprove that $$f(a\cap b) = f(a)\cap f(b)$$ for every function $f$ and every subsets $a$ and $b$ of the domain of $f$. Would you be able to proceed to prove or disprove that? That is part of what you need to do here, to resolve item 2.

Then proceed with the other three items.

MJD
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$a$ and $b$ are both sets ($a,b \subseteq X$), but the function maps elements of $X$ to elements of $Y$. So $f(a) = \{f(x) \mid x \in a\}$ is the image of the subset $a$ under $f$. You should now calculate $f(a \cap b)$, i.e. the image of the subset $a \cap b$ under $f$.

inyono
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Let $x \in A \cap B$ with $A,B \subset X$, then $f(x)\subset f(A)$ (because $x \in A$) and $f(x)\in f(B)$. So $f(x)\in f(A)\cap f(B)$. It follows that $f(A\cap B)\subset f(A)\cap f(B)$. Can you show an example/counter example to determine if it is a proper subset or not?

Surb
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