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I have to prove:

An infinite subset of a denumerable set is denumerable.

I understand this has been asked before and I did take the time to read what was said there, but I do not understand still.

I have to prove this using other theorems about denumerable or countable sets, nothing too complex for this proof.

Sam
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  • There is some ambiguity in what you mean by countable. See the first few paragraphs at https://en.wikipedia.org/wiki/Countable_set – parsiad Dec 07 '16 at 22:57
  • @parsiad For me a set is countable if it is finite or denumerable. – Sam Dec 07 '16 at 22:59
  • So you know that subsets of countable sets are countable? – Lukas Betz Dec 07 '16 at 23:12
  • @LeBtz yes I did mention that. We did learn that theorem. – Sam Dec 07 '16 at 23:41
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    Then I don't understand the difficulty. An infinite subset of a denumerable set is an infinite subset of a countable set and therefore countable itself. Now it can't be finite because it's infinite by assumption so it must be denumerable. – Lukas Betz Dec 07 '16 at 23:46
  • @LeBtz How are you getting that an infinite subset of a denumerable set is an infinite subset of a countable? Isn't countable and denumerable different? – Sam Dec 07 '16 at 23:54
  • As far as I know, denumerable set and countable set are synonyms. – DanielWainfleet Dec 08 '16 at 02:26
  • @user254665 are they really? I thought there was a difference? Now I am confused... – Sam Dec 08 '16 at 02:51
  • Your own definition of countable says that each denumerable set is countable by definition. – Lukas Betz Dec 08 '16 at 09:40
  • @LeBtz I just understood what you said, it just clicked. I get it now. Thank You! – Sam Dec 08 '16 at 22:25

2 Answers2

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Suppose that the subset is not a countable infinite set, so if you prove that your subset $A$ must be more than countable in order to be infinite you have done, because then it can't be a subset of you denumerble set. Suppose that there exists an infinite set $B$ whose cardinality is less than the cardinality of the natural numbers that is infinite; so there exists an injective function $f: B \rightarrow \mathbb N$: its image is infinite by hypothesis, and it is a subset of the natural numbers so it has a minimum $x_0$: associate this minimum with $0$ and then consider the set $f(B)\setminus x_0$ and iterate considerating the new minimum $x_1$; your process never ends because $f(B)$ is infinite. So we have shown that there is a bijection between $B$ and $\mathbb N$, that is what we wanted to prove.

Bargabbiati
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One of the many possible proofs:

If a set $A$ is denumerable, then there is a bijection $f:A\rightarrow \mathbb N$

So if $B\subseteq A$,the restriction of $f$ to $B$, which is $f:B\rightarrow \mathbb N$ is $1:1$

On the other side, if $B$ is infinite, we can construct an $1:1$ function $g:\mathbb N\rightarrow B$ by defining $f(0)=x_0\in B$, then $f(1)=x_1\in B-\{x_0\}$, and so on by induction.

So by Schröder–Bernstein there is a bijection between $B$ and $\mathbb N$, which makes $B$ denumerable.

User125687
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