Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
—James R. Munkres
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))
and )\subset%20B_0)
. The first inclusion is an equality if 
is injective and the second if surjective.
Proof. )
in the domain of 
such that )=a_0.)
Thus .
Let ))
, then there exists 
in the range of such that =a')
and )
for some 
. Thus =f(a))
. Suppose is injective, then we have 
and hence 
. Therefore, )%20\subset%20A_0)
.
))
, there exists )
such that =b)
, =a)
for some 
. Thus =b_0)
and =b_0\in%20B_0)
. Suppose is surjective. For every 
, there exists 
in the domain of such that =b_0)
. Thus \subset%20f^{-1}(B_0))
and %20%20\subset%20f(f^{-1}(B_0)))
. Therefore, ))
.
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for 
Show that preserves inclusions, unions, intersections, and differences of sets:
(a) \subset%20f^{-1}(B_1))
.
(b) =f^{-1}(B_0)\cup%20f^{-1}(B_1))
.
(c) =f^{-1}(B_0)\cap%20f^{-1}(B_1))
.
(d) =f^{-1}(B_0)-f^{-1}(B_1))
.
Show that preserves inclusions and unions only:
(e) \subset%20f(A_1))
.
(f) =f(A_0)\cup%20f(A_1))
.
(g) %20\subset%20f(A_0)\cap%20f(A_1))
; show that equality holds if is injective.
(h) %20\supset%20f(A_0)-%20f(A_1))
; show that equality holds if is injective.
Proof. (a) %20=%20\{a|%20f(a)%20\in%20B_0%20\}\subset%20\{a|f(a)\in%20B_{1}\}%20=%20f^{-1}(B_1).)
(b) =%20\{a|%20f(a)%20\in%20B_0%20\cup%20B_1\}%20=%20\{a|%20f(a)%20\in%20B_0\}%20\cup%20\{a|%20f(a)%20\in%20B_1\}%20=f^{-1}(B_0)\cup%20f^{-1}(B_1))
(c) =%20\{a|%20f(a)%20\in%20B_0%20\cap%20B_1\}%20=%20\{a|%20f(a)%20\in%20B_0\}%20\cap%20\{a|%20f(a)%20\in%20B_1\}%20=f^{-1}(B_0)\cap%20f^{-1}(B_1))
(d) =%20\{a|%20f(a)%20\in%20B_0%20-%20B_1\}%20=%20\{a|%20f(a)%20\in%20B_0\}%20-%20\{a|%20f(a)%20\in%20B_1\}%20=f^{-1}(B_0)-%20f^{-1}(B_1))
(e) For every )
, there exists 
such that . But 
, so )
.
(f) =\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_0%20\cup%20A_1\}=\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_0\}%20\cup%20\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_1\}=f(A_0)\cup%20f(A_1))
(g) %20=\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_0%20\cap%20A_1\}\subset%20\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_0\}%20=%20f(A_0))
. Similarly %20\subset%20%20f(A_1))
. Thus .
If \cap%20f(A_1))
, then 
such that =f(a_1)=b)
. Since is injective, 
. Thus 
and \in%20f(A_0\cap%20A_1))
.
(h) -%20f(A_1)%20=%20\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_0\}%20-%20\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_1\})
. If -%20f(A_1))
then )
for some 
but there is no 
such that =b)
. Thus 
. Thus 
. Thus %20=%20\{b|%20b=f(a)%20,%20\exists%20a%20\in%20A_0-A_1\})
. Hence .
If . Thus 
but 
. 
, \ne%20f(a)=b)
by injective assumption. Thus )
. Thus -f(A_1))
. Hence %20\subset%20f(A_0)-%20f(A_1))
.
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Show that (b), (c), (f), and (g) of Exercise 2 hold for arbitrary unions and intersections.
and 
are two collection of sets.
(b’) =\cup_{B\in%20\mathcal{B}}%20f^{-1}(B))
(c’) =\cap_{B\in%20\mathcal{B}}%20f^{-1}(B))
(f’) =\cup_{A\in%20\mathcal{A}}%20f(A))
(g’) %20\subset%20\cap_{A\in%20\mathcal{A}}%20f(A))
; show that equality holds if is injective.
Proof. (b’), (c’), and (f’) are just by definitions of union, intersection, and preimage.
(g’) If )
then there exists at least an 
such that . Then 
is in each set of 
. Thus for each there exists 
such that \in%20f(A))
. Thus )
. Hence .
If then 
is in each ,%20\forall%20A\in%20\mathcal{A})
. Then for each 
there exists such that . By injective assumption we must have these in each 
are the same. Thus and \in%20f(\cap_{A\in%20\mathcal{A}}%20%20A))
. Hence %20\supset%20\cap_{A\in%20\mathcal{A}}%20f(A))
.