Let f be a function from the set A to the set B. Let S and T be subsets of A. Show that \$f(S cup T) = f(S) cup f(T)\$

\$yin f(Scup T) ightarrow exists s in S cup T ;such ;that ;y= f(x) \$

\$if ; x in S ;then ;y= f(x) in f(S) subset f(S) cup f(T) ightarrow y in f(S) cup f(T)\$

\$if ; x otin S ;then ;x in T ; and ;;y = f(x) in f(T) subset f(S) cup f(T) ightarrow y in f(S) cup f(T)\$

\$it ;follows ;that ; f(S cup T) subset f(S) cup f(T)\$

\$Since ;S,T subset S cup T, f(S), f(T) subset f(S cup T) ightarrow f(S) cup f(T) subset f(S cup T)\$

\$By ;above, ;we ;have ;f(S cap T) =f(S) cap f(T)\$

I cannot understand why I need to consider \$x in S\$ and \$x otin S\$? and Why the \$f(S)\$ is the subset of \$f(S) cup f(T)\$? and Why in the step three the \$f(T)\$ is also the subset of \$f(S) cup f(T)\$?

elementary-set-theory functions relations
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Folshort
edited Dec 16 "12 at 7:03 Martin Sleziak
asked Dec 16 "12 at 4:53 SamuelSamuel
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If you only consider \$x in S\$, then you haven"t considered all the elements in \$S cup T\$ because there might be elements in \$T\$ that are not in \$S\$.

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And \$f(T)\$ is a subset of \$f(S) cup f(T)\$ because the union of two sets will have each of the two sets as the subset.

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Folshort
answered Dec 16 "12 at 4:58 Joe Z.Joe Z.
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