Be f : A → B a function and X,Y ⊂ A. Show that: any thoughts? thanks

a) #f (X U Y) = f(X) U f(Y)#

b)If #f# is injective, then #f(X-Y) = f(X) - f(Y)#

1 Answer
Jim H
Jan 3, 2018

See below.

Explanation:

a)
Let #b# be an element of #f(X uu Y)#.
Then #b = f(a)# for some #a in XuuY#.
#a in X rArr f(a) = b in f(X)#
#a in Y rArr f(a) = b in f(Y)#
#a in X or a in Y#
Therefore #f(a) = b in f(X)uuf(Y)#
This shows that #f(XuuY) sube f(X)uuf(Y)#

For #binf(X)uuf(Y)#, we have
#b=f(a)# for some #a in X# or for some #a in Y#
In either case #b = f(a)# for some #a in XuuY# so
#b in f(XuuY)#
This shows that # f(X)uuf(Y) sube f(XuuY)#.

We conclude that #f(XuuY) = f(X)uuf(Y)#

b) is similar.
First show that #f(X - Y) sube f(X)-f(Y)#
Let #b# be an element of #f(X - Y)#.
Then #b = f(a)# for some #a in X and a !inY#.
So #b in f(X) and b!in f(Y)# because #f# is injective.
Hence, #b in f(X)-f(Y)#

Now show that #f(X) - f(Y) sube f(X-Y)#
Let #b# be an element of #f(X) - f(Y)#
Then #b in f(X) and b !in f(Y)#
So #b = f(a)# for some #a in X#
If #a in Y# then #b=f(a) in f(Y)# contrary to #b !in f(Y)#
Hence #a !in Y#
Therefore #a in X-Y# and
#b = f(a) in f(X-Y)#

We can conclude that #f(X - Y) = f(X)-f(Y)#

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