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For each of the following sets, determine if it is finite, countably infinite, or uncountable. You need not prove your answer is correct, but you should give a reason for your response.

  1. For some $n\in\mathbb{N}$, $\{(a_1, a_2, \dots, a_n)\in\mathbb{N}^n\ | \ a_1+a_2+\dots+a_n<5000\}$
  2. $\{S\in\mathcal{P}(\mathbb{N})\ | \ \mathbb{N}\backslash S\hbox{ is finite}\}$
  3. $\{x\in \mathbb{R}\ | \ \hbox{there is a decimal expansion of $x$ with only even digits}\}$

For #1, I said that it is finite. Since every $a_i \in \mathbb{N}$, every $a_i \leq 5000$. Combinations of a finite set of numbers must also be finite.

For #2, I said that it is countably infinite. To make $\mathbb{N}\backslash S$ finite, $S$ would have to be an infinite subset of $\mathbb{N}$, of which are infinite. Therefore the initial set is infinite. Since $\mathbb{N}$ is countably infinite, we can reason that the initial set is also countably infinite.

For #3, I said that it is uncountable. An argument can be made with Cantor's diagonalization method to show that a list of real numbers with decimal expansions with only even digits can never be complete and the set is therefore uncountable.

Do my arguments make sense/ are my conclusions correct?

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The each claim is correct. I would express the first argument as "since the set's cardinality is at most $5000^n,$ then the set has finite cardinality," or something similar.

Your argument for the second is hard to make sense of, unfortunately. I would say, rather, that $\Bbb N$ has countably-infinitely-many finite subsets (as a countable union of countable sets is countable), so it has countably-infinitely-many subsets with finite complements.

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  • $\begingroup$ Let $\mathscr S$ be the set defined in $2$. Then there is a $1-1$ correspondence between finite subsets of $\Bbb N$ and elements of $\mathscr S$. Since there are countably many finite subsets of $\Bbb N$, then there are also countably many elements of $\mathscr S$. This is a slightly more precise way of saying the second paragraph in this answer. $\endgroup$ – Robert Shore Apr 9 at 2:48
  • $\begingroup$ @Robert - True. $\endgroup$ – Cameron Buie Apr 9 at 2:50

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