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Which of the following sets are finite? countably infinite? uncountable? (Be careful -- don't apply theorems for finite sets to infinite sets and don't apply theorems for countable sets to uncountable sets!) Give reasons for your answers for each of the following:

(a) $\{1/n: n \in \mathbb{Z} \setminus \{ 0 \} \}$;

(b) $\mathbb{R} \setminus \mathbb{N}$

(c) $\{x \in \mathbb{N}: \lvert x-7\rvert > \lvert x \rvert \}$;

(d) $2\mathbb{Z} \times 3 \mathbb{Z}$

For (a), I think this set=$\mathbb{Q}$ which we know to be countably infinite? (b) is infinite, but I'm not sure how to tell if it is countable. (c) I want to say is countably infinite just by thinking about the set. (d) Seems countably infinite as well since $\mathbb{Z}$ is countably infinite?

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    $\begingroup$ For some reason there were too many dollar signs in your LaTeX code (that I have edited), most of them instead of blank spaces. If you want to insert LaTeX formulae, just insert them between two matching dollar signs at the ends, not more. $\endgroup$ – Alex M. Apr 17 '15 at 16:48
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For (a): this is not $\mathbb{Q}$, since it only consists of fractions having numerator $1$, not of all fractions. Ans yes, this is countable, since it is in bijection with $\mathbb{Z} \setminus \{0\}$, which itself is countable.

(b): clearly not countable. If it had been countable, since $\mathbb{N}$ is also countable and the union of 2 countable sets is countable, then $\mathbb{N} \cup (\mathbb{R} \setminus \mathbb{N}) = \mathbb{R}$ would also be countable, which is false.

(c): if $x \geq 7$, then the inequality becomes $x-7 > x$, which cannot have any solution. Therefore, it may consist of at most $\{0,1,2,3,4,5,6\}$, which is a finite set. (In fact, it consists only of $\{0,1,2,3\}$.)

(d): correct; the product of 2 countable sets is itself countable.

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