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There is an assignment which I need to complete for school and I have absolutely no clue on how to solve the following question.

If A and B are sets and $f: A \rightarrow B$, then for any subset $S$ of $A$ we define

$f(S) = \{ b \in B:b = f(a)$ for some $a \in S\}.$

Similarly, for any subset $T$ of $B$ we define the pre-image of $T$ as

$f^{-1} (T) = \{a \in A : f (a) \in T\}$

Note that $f^{-1} (T)$ is well defined even if $f$ does not have an inverse!

Now let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as $f(x) = x^2$. Let $S_1$ denote the closed interval $[-2,1]$, that is all $x\in \mathbb{R}$ that satisfy $-2\le x\le1$, and let $S_2$ be the open interval $(-1,2)$, that is all $x\in \mathbb{R}$ that satisfy $-1 < x<2$. Also let $T_1 = S_1$ and $T_2 = S_2$.

Determine

$f(S_1\cup S_2)$, $f(S_1)\cup f(S_2)$, $f(S_1 \cap S_2)$, $f(S_1) \cap f(S_2)$,

and

$f^{-1} (T_1 \cup T_2)$, $f^{-1} (T_1) \cup f^{-1}(T_2)$, $f^{-1} (T_1 \cap T_2)$, and $f^{-1} (T_1) \cap f^{-1}(T_2)$

Jake0991
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2 Answers2

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Hint: For $f(S_1\cup S_2)$, first determine what $S_1\cup S_2$ is. Do you see that it is $[-2,2)$? Now, what is $f([-2,2))$ (that is, what are the squares of each number in $[-2,2)$?)

For $f(S_1)\cup f(S_2)$, the problem asks for the union of $f(S_1)$ and $f(S_2)$, so first you need to compute these two things, then take their union.

The other parts are similar.

rogerl
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As $\forall x\in \mathbb R\;\;\; x^2\geq 0$,

and $S_2\subset S_1$,

$$f(S_1 \cup S_2)=f(S_1)=[0,4]$$

$$f(S_1)\cup f(S_2)=f(S_1)$$

$$f(S_1\cap S_2)=f(S_2)=[0,4)$$

$$f(S_1)\cap f(S_2)=f(S_2)$$

observe that

$f^{-1}((-\infty,a^2])=[-a,a]$ and $T_2\subset T_1$.

$$f^{-1}(T_1)=[-1,1]$$

$$f^{-1}(T_2)=(-1,1)$$ you can finish.

hamam_Abdallah
  • 62,951