Consider the following argument for showing that |R| = |P(N)|. By the Cantor-Bernstein-Schröder theorem (page 17, following Definition 0.3.28), it suffices to find an injection from P(N) into R and also an injection from R into P(N). When S is a set of natural numbers, create a real number is in the following way. Start from the decimal 0.123456789101112... obtained by writing down the natural numbers in their standard order, and for each element n of the set S, replace the instance of ŉ in this decimal by a string of zeros (one for each digit of n). For example, if S is the set of even natural numbers, then as is the decimal 0.10305070900110013.... The function S→ as is an injection from P(N) into R. According to the discussion on page 38 (after Proposition 1.4.1), all intervals have the same cardinality, so instead of producing an injection from R into P(N), it suffices to find an injection from (0, 1) into P(N). When a is a real number between 0 and 1, write a as a nonterminating decimal. Associate to x a set S₂ of natural numbers in the following way. For each positive integer n, if the nth decimal digit of x is not 0, form a natural number by appending n zeros to this digit, and put this natural number into the set S. For example, if x = 0.409 ..., then S₂ contains the natural numbers 40 and 9000 and so forth. The function → S₂ is an injection from (0, 1) into P(N). True or false: This argument proves the claim on page 39 just above Theorem 1.4.2 that |R| = |P(N)|.

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter8: Polynomials

Section8.4: Zeros Of A Polynomial

Problem 18E: Show that the converse of Eisenstein’s Irreducibility Criterion is not true by finding an...

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Question

Consider the following argument for showing that |R| = |P(N)|.
By the Cantor-Bernstein-Schröder theorem (page 17, following
Definition 0.3.28), it suffices to find an injection from P(N) into R and
also an injection from R into P(N).
When S is a set of natural numbers, create a real number is in the
following way. Start from the decimal 0.123456789101112... obtained
by writing down the natural numbers in their standard order, and for each
element n of the set S, replace the instance of n in this decimal by a
string of zeros (one for each digit of n). For example, if S is the set of
even natural numbers, then as is the decimal
0.10305070900110013.... The function S→s is an injection from
P(N) into R.
According to the discussion on page 38 (after Proposition 1.4.1), all
intervals have the same cardinality, so instead of producing an injection
from R into P(N), it suffices to find an injection from (0, 1)
into P(N). When x is a real number between 0 and 1, write x as a
nonterminating decimal. Associate to x a set S of natural numbers in
the following way. For each positive integer n, if the nth decimal digit
of x is not 0, form a natural number by appending n zeros to this digit,
and put this natural number into the set S. For example, if
x = 0.409 ..., then S, contains the natural numbers 40 and 9000 and
so forth. The function → S is an injection from (0, 1) into P(N).
True or false: This argument proves the claim on page 39 just above
Theorem 1.4.2 that |R| = |(N).

Expert Solution

Step 1

As per the question, we will be verifying the argument $|ℝ| = |P (ℕ)|$ .

Let us summarize the given information.

By the Cantor-Bernstein-Schroder theorem, it suffices to find an injection from $P (ℕ)$ into $ℝ$ and also an injection from $ℝ$ into $P (ℕ)$ .

$(\Rightarrow)$

Let us assume $S$ is a set of natural numbers.

So, $S \in P (ℕ)$ .

Now, $x_{S}$ is created in the following way.

Start from the decimal $0.123456789101112 . . .$ obtained by writing down the natural numbers in their standard order, and for each element $n$ of the set $S$ , replace the instance of $n$ in this decimal by a string of zeros (one for each digit of $n$ ).