1.4.E: Problems on Countable and Uncountable Sets (Exercises)
- Page ID
- 22253
Prove that if \(A\) is countable but \(B\) is not, then \(B-A\) is uncountable.
[Hint: If \(B-A\) were countable, so would be
\[
(B-A) \cup A \supseteq B . \quad(\mathrm{Why?})
\]
Use Corollary \(1 . ]\)
Let \(f\) be a mapping, and \(A \subseteq D_{f} .\) Prove that
(i) if \(A\) is countable, so is \(f[A]\) ;
(ii) if \(f\) is one to one and \(A\) is uncountable, so is \(f[A]\).
\(\left[\text { Hints: }\left(\text { i) If } A=\left\{u_{n}\right\}, \text { then }\right.\right.\)
\[
f[A]=\left\{f\left(u_{1}\right), f\left(u_{2}\right), \ldots, f\left(u_{n}\right), \ldots\right\}
\]
(ii) If \(f[A]\) were countable, so would be \(f^{-1}[f[A]],\) by (i). Verify that
\[
f^{-1}[f[A]]=A
\]
here; cf. Problem 7 in §§4-7.]
Let \(a, b\) be real numbers \((a<b) .\) Define a map \(f\) on \([0,1)\) by
\[
f(x)=a+x(b-a).
\]
Show that \(f\) is one to one and onto the interval \([a, b)=\{x | a \leq x<b\}\). From Problem \(2,\) deduce that \([a, b)\) is uncountable. Hence, by Problem \(1,\) so \(i s(a, b)=\{x | a<x<b\}\).
Show that between any real numbers \(a, b(a<b)\) there are uncountably many irrationals, i.e., numbers that are not rational.
[Hint: By Corollary 3 and Problems 1 and \(3,\) the set \((a, b)-R\) is uncountable. Explain in detail.
Show that every infinite set \(A\) contains a countably infinite set, i.e., an infinite sequence of distinct terms.
[Hint: Fix any \(a_{1} \in A ; A\) cannot consist of \(a_{1}\) alone, so there is another element
\[
a_{2} \in A-\left\{a_{1}\right\} . \quad(\mathrm{Why} ?)
\]
Again, \(A \neq\left\{a_{1}, a_{2}\right\},\) so there is an \(a_{3} \in A-\left\{a_{1}, a_{2}\right\} .\) (Why?) Continue thusly ad infinitum to obtain the required sequence \(\left\{a_{n}\right\} .\) Why are all \(a_{n}\) distinct? \(]\)
From Problem \(5,\) prove that if \(A\) is infinite, there is a map \(f : A \rightarrow A\) that is one to one but not onto \(A .\)
[Hint: With \(a_{n}\) as in Problem \(5,\) define \(f\left(a_{n}\right)=a_{n+1} .\) If, however, \(x\) is none of the \(a_{n},\) put \(f(x)=x\). Observe that \(f(x)=a_{1}\) is never true, so \(f\) is not onto \(A .\) Show, however, that \(f\) is one to one.
Conversely (cf. Problem 6), prove that if there is a map \(f : A \rightarrow A\) that is one to one but not onto \(A,\) then \(A\) contains an infinite sequence \(\left\{a_{n}\right\}\) of distinct terms.
[Hint: As \(f\) is not onto \(A,\) there is \(a_{1} \in A\) such that \(a_{1} \notin f[A] .\) (Why?) Fix \(a_{1}\) and define
\[
a_{2}=f\left(a_{1}\right), a_{3}=f\left(a_{2}\right), \ldots, a_{n+1}=f\left(a_{n}\right), \ldots \text { ad infinitum. }
\]
To prove distinctness, show that each \(a_{n}\) is distinct from all \(a_{m}\) with \(m>n .\) For \(a_{1},\) this is true since \(a_{1} \notin f[A],\) whereas \(a_{m} \in f[A](m>1) .\) Then proceed inductively.]