Question

Let $\mathit{f}\mathbf{:}\mathit{B}\mathbf{\to }\mathit{C}$be the function and let be subsets of .

(a) Prove that $\mathit{f}\mathbf{\left(}\mathit{S}\mathit{U}\mathit{T}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathit{U}\mathit{f}\mathbf{\left(}\mathit{T}\mathbf{\right)}$.

(b) Prove thatlocalid="1659164248456" $\mathit{f}\mathbf{(}\mathit{S}\mathit{I}\mathbf{}\mathit{T}\mathbf{)}\underline{\mathbf{\subset }}\mathit{f}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathit{I}\mathit{f}\mathbf{\left(}\mathit{T}\mathbf{\right)}$

(c) Give an example where$\mathit{f}\mathbf{(}\mathit{S}\mathit{I}\mathbf{}\mathit{T}\mathbf{)}\mathbf{\ne }\mathit{f}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathit{I}\mathbf{}\mathit{f}\mathbf{\left(}\mathit{T}\mathbf{\right)}$

Short answer

(a) It is proved that $f\left(SUT\right)=f\left(S\right)Uf\left(T\right)$.

(b) It is proved that $f(SIT)\underline{\subset }f\left(S\right)If\left(T\right)$.

(c) It is proved that localid="1659166606129" $f(SIT)\ne f\left(S\right)If\left(T\right)$.

Step-by-step solution

(a) Step 1: Showing that f(SU T)≠f(S)U f(T)

As it is given that $f:B\to C$be a function and let S, Tbe subsets of .Now, assume that there is a point c which is $f\left(b\right)=c$for some $b\in \left(SUT\right)$. Either $b\in S$and f (b) belongs in f (S).

Again, consider that there is a point c such that $c\in \left(f\right(S\left)Uf\right(T\left)\right)$. It means that c is either a point in f (S)or cis a point in f (T) meansf has a pre-image either in or .

Thus, there is an element of $f\left(SUT\right)$. It means that $f\left(S\right)Uf\left(T\underline{\subset }f\right(SUT\left)\right)$.

But if role="math" localid="1659165575937" $f\left(SUT\right)\underline{\subset }f\left(S\right)Uf\left(T\right)\mathrm{and}f\left(S\right)Uf\left(T\underline{\subset }f\right(SUT\left)\right)$the written as follows:

$f\left(SUT\right)=f\left(S\right)Uf\left(T\right)$

Hence, it is proved that $f\left(SUT\right)=f\left(S\right)Uf\left(T\right)$.

(b) Step 2: Showing that f(SIT)⊂f(S)Uf(T) 

Assume that there is a pointc which is $c\in f(SIT)$then $c\in f\left(b\right),\exists b\in (SIT)$for some role="math" localid="1659165892700" $b\in (SUT)$. Either $b\in S$and f (b) belongs in $f\left(S\right)$.

It is implies that $b\in S$and $b\in T$so it is known that $c\in f\left(S\right)$and $c\in f\left(T\right)$.

Thus, c is the element of $f\left(S\right)If\left(T\right)$.

Hence, it is proved that $f(SIT)\underline{\subset }f\left(S\right)If\left(T\right)$.

(c) Step 3: Showing that f(SI T)≠f(S)I f(T)

As $f(SIT)\subset f\left(S\right)If\left(T\right)$now set is found to make this fail whenever fis not injective or it is said that $m\ne n$but $f\left(m\right)=f\left(n\right)$. Take $S=\left\{m\right\},T=\left\{b\right\}$, then:

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

As now, $f(SIT)\subset f\left(S\right)If\left(T\right)$let $y\in f(SIT)$then y = f (x) for some $y\in SIT$.

Since, xis in both S and T, y is in both f (S)and f(T).

It is implies that $b\in S$and $b\in T$so it is known that $c\in f\left(S\right)$and $c\in f\left(T\right)$.

Thus, $y\in f\left(S\right)If\left(T\right)$.

Hence, it is proved that $f(SIT)\ne f\left(S\right)If\left(T\right)$.