Section 2.8: Problem 6 Solution

Working problems is a crucial part of learning mathematics. No one can learn... 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 $^{*}\mathfrak{R}$ . Munkres

(a) Show that $^{∗}\mathbb{Q}$ has cardinality at least $2^{\aleph_{0}}$ , where $\mathbb{Q}$ is the set of rational numbers. Suggestion: Use Exercise 1.

(b) Show that $^{∗}\mathbb{N}$ has cardinality at least $2^{\aleph_{0}}$ .

(a) According to Exercise 1, for every $r\in\mathbb{R}$ , there is $q_{r}\in{}^{*}\mathbb{Q}$ such that $r\simeq q_{r}$ , but if $r\neq r'$ , then $q_{r}\simeq r\not\simeq r'\simeq q_{r'}$ , i.e. $q_{r}\neq q_{r'}$ , implying that $|\mathbb{R}|=2^{\aleph_{0}}\le|{}^{*}\mathbb{Q}|$ .

(b) There is a one-to-one function $S:\mathbb{Q}\rightarrow\mathbb{N}$ , therefore, according to Exercise 2(a), $^{*}S:{}^{*}\mathbb{Q}\rightarrow{}^{*}\mathbb{R}$ . In fact, $^{*}S:{}^{*}\mathbb{Q}\rightarrow{}^{*}\mathbb{N}$ , as the sentence $\forall v_{1}\forall v_{2}(P_{S}v_{1}v_{2}\rightarrow P_{N}v_{2})$ , where $N$ is the relation that defines $\mathbb{N}$ in $\mathbb{R}$ , is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ . Moreover, since the sentence $\forall v_{1}\forall v_{2}\forall v_{3}(P_{S}v_{1}v_{3}\rightarrow P_{S}v_{2}v_{3}\rightarrow v_{1}=v_{2})$ is true in $\mathfrak{R}$ , and, hence, in $^{*}\mathfrak{R}$ , $^{*}S$ is injective (one-to-one), and, according to (a), $2^{\aleph_{0}}\le|{}^{*}\mathbb{Q}|\le|{}^{*}\mathbb{N}|$ .

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Section 2.8: Problem 6 Solution

Section 2.8: Problem 6 Solution