Question

Prove that $\mathit{f}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{C}$is injective if and only if$\mathit{f}\left(SIT\right)\mathbf{=}\mathit{f}\left(S\right)\mathit{I}\mathbf{}\mathit{f}\left(T\right)$ for every pair of subsets S,T of .

Short answer

Thus, the given statement is proved.

Step-by-step solution

The statement f : B→C is injective

Consider the provided statement to prove that $f:B\to C$injective if and only if $f(\mathrm{S}\mathrm{I}\mathrm{T})=f\left(\mathrm{S}\right)If\left(\mathrm{T}\right)$ for every pair of the subsets S.T $\underline{\subset }B$

Now, it is to be proved that if $f:B\to C$is injective, let $y\in f\left(SIT\right)$. Therefore,$y\in f\left(S\right)$thus, $y=f\left(s\right)$for some $s\in S$and $y\in f\left(T\right)$therefore, $y=f\left(t\right)$for some $t\in T$.

But as it is given that, f is injective so f ( s ) = f ( t ) and implies that s = t.

Hence, s = t$\in f\left(S\right)If\left(T\right)$

Thus, write as follows:

$$\begin{gathered} y=f\left(s\right) \\ =f\left(t\right)\in f\left(SIT\right) \end{gathered}$$

Therefore,$f\left(SIT\right)=f\left(S\right)If\left(T\right)$ is proved.

The statement f (S I T)= f (S)I f (T)is injective

Now, the second implication is proved as below:

Now, it is considered that f(S I T) = f (S) I f (T) for all subsets S , T$\underline{\subset }B$ .

Let role="math" localid="1659161302739" $x_1,x_2\in B$then $S=\left\{x_1\right\},T=\left\{x_2\right\}$. If$x_1\ne x_2$ then$SIT=\varphi$ therefore,

$$\begin{gathered} f\left(SIT\right)=f\left(\varphi \right) \\ =\varphi \end{gathered}$$

That is, f ( S )I f (T)=$\varphi$ therefore,$f\left(x_1\right)\ne \left(x_2\right)$ so, f is injective.

Hence, the statement is proved.