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EDIT: The original question is: Let $N$ be a countable set and $f$ a function that maps $N$ onto a set $M$. Prove that $M$ is countable.

Consider a countable set $N$ which the function $f$ maps [on]to a set $M$. I want to prove that the set $M$ also is countable, if $N$ is. The thing I am unsure about, is if $f(N)=M$?

Here is my proof, though:

Proof. Let $N \subset X$ be a countable set and $f:X \rightarrow Y$.

Therefore, $N\cup f(N)=\left \{ x_1, f(x_1), x_2, f(x_2),...,x_n,f(x_n) \right \}$ which can be expressed as

$N\cup f(N)=\left \{x_1,f(x_1) \right \}\cup\left \{x_2,f(x_2) \right \}\cdots \cup\left \{x_n,f(x_n) \right \}$.

Theorem (*), picked from Set Theory and Matrices by I. Kaplansky : A countable union of countable sets is countable.

Since we have a union of $n$ sets in which there is two elements, $N\cup f(N)$ is therefore countable according to (*).

Since $N$ is defined as countable and $N\cup f(N)$ is countable, thus $f(N)$ is countable.

The question is though, is $f(N)$ referred as "the same thing" as $M$? If the purpose is proving that an arbitrary set $M$ is countable, I am totally lost.

Does anyone here have any suggestions?

Andrés E. Caicedo
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  • Take $N = {1}$ and $M = \mathbb{R}$, then define the map $f : N \to M$ by $f(1) = 47$. $N$ is countable, but $M$ is not. In other words, the range of a function needn't be the entire codomain. If the result is to be true, then $f$ needs to be surjective, I suppose (meaning that $f(N) = M$). – Xander Henderson Feb 26 '19 at 21:31
  • How does one prove that $M$ is countable though? –  Feb 26 '19 at 21:35
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    Well, obviously $f(N) = {f(n)|n \in \mathbb N$ is countable as $f$ is a surjective function onto $f(N)$. However if $f$ in not surjective to $M$ and $f(N) \subsetneq M$ that's not a damn thing you can say about the countability of $M$. For examaple. Let $i: \mathbb N \to \mathbb R$ via $i(n) = n$. Then we have a surjection (actually a bijection) from $\mathbb N \to i(\mathbb N) = \mathbb N\subsetneq \mathbb R$. So $i(\mathbb N) = \mathbb N$ is countable. But what can we say about $\mathbb R$ being countable? Not a damned thing. – fleablood Feb 26 '19 at 21:38
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    "How does one prove that M is countable though?" You don't have a hope in hell of being able to do that. $M$ can be any set that has $f(N)$ as a proper subset and that can have any cardinality at least as high as $\mathbb N$. and Xander just gave you an example where $M$ is NOT countable. – fleablood Feb 26 '19 at 21:41
  • "The original question is: Let N be a countable set and f a function that maps N *!!!onto!!!* a set M". If $f$ is *!!!onto!!!* then $f(N) = M$. That is what onto means. – fleablood Feb 26 '19 at 22:05
  • In other news, prepositions matter. – Xander Henderson Feb 27 '19 at 02:02

2 Answers2

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An essential missing condition in your stated problem is that $M$ is defined by the set of points reachable as $f(n)$ for some $n\in N$. That is, $f:N\rightarrow M$ is surjective. You could say that the statement "$N$ maps to $M$ implies that condition, but it is nice to be clear about it.

With that understood, the outline of the proof that was probably wanted is that since $N$ is countable, there is some surjective counting function $k: \Bbb{N} \rightarrow N$. Now consider the map $g : \Bbb{N} \rightarrow M$ such that for all $i \in \Bbb{N}$, $$ g(i) = k(f(i)) $$ Because $k$ is surjective, the map $g$ applies $f(n)$ to every member of $N$, and then because $f$ is surjective onto $M$, that covers every member of $M$.

This proof is not rigorous but can be made rigorous without introducing any new ideas or theorems.

The proof you presented has the weakness that it uses a stronger theorem than is really necessary (countable union of countable sets), but it isn't actually wrong.

Mark Fischler
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$N$ is countable.

That means there is an $A\subset \mathbb N$ and an $j$ so that $j:A \to N$ is bijective.

We are told $f: N\to M$ is !ONTO!. Because $f$ is ONTO that means $f(N) = M$. We were given that.

So $h=f\circ j: A\subset \mathbb N \to N \to M$ is onto. We can use $h:A \to M$ directly and not have to deal with the intermediate set $N$ anymore. That'll make things easier.

But we don't know that $h$ is injective; just that it is onto.

But let $B \subset A \subset \mathbb N$ be defined like this: for each $m \in M$ there are one or more $n_i \in A$ so that $f(n_i)=m$. Let the least of those $n_i$ be in $B$ but none of the others.

Let $h': B \to M$ via $h'(n) = h(n)$ for each $n \in B$.

$h'$ is a bijection.

Pf: $h'$ is onto: For every $m \in M$ there are $n_i\in A$ so that $h(n_i) = m$ and if $n$ is the least of the $n_i$ then $n\in B$ and $h'(n) =m$.

$h'$ is one to one: If $h'(n) = h'(k) = m$ then $n,k \in A$ and $h(n) = h(k) = m$. But if $n\ne k$ then only the smaller of them were in $B$. So $n$ and $k$ can't be different as they are in $B$. So $n=k$.

So $h'$ is a bijection from $B\subset \mathbb N$ to $M$ so $M$ is countable.

[Intuitively if $f: A \to B$ is onto then the cardinality of $B$ is "at most" the cardinality of $A$ because every thing in $B$ is mapped from an element in $A$.]

fleablood
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