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
edited Dec 16 "12 at 7:03
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asked Dec 16 "12 at 4:53
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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.
answered Dec 16 "12 at 4:58
Joe Z.Joe Z.
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