Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have an example is like this

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

$answer:$

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

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

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

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

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

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

I cannot understand why I need to consider $x \in S$ and $x \notin S$? 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 $f(S) \cup f(T)$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $S$.

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.