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My problem reads:

Prove that if there is a function $f\colon A\rightarrow \mathbb{N}$ that is one-to-one, then $A$ is countable

Assuming $\mathbb{N}$ to be the set of natural numbers.

I am not too sure how to go about proving this. Would I need the definition of denumerable in this case? or can I use the cardinality of A < or equal to N?

Sam
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  • If $f:A \to N$ is one-to-one, then $A$ is in bijective correspondence with the range of $f$, which is a subset of $n$, say $S={n_1,n_2, \ldots}$ (If $S$ is finite, then so is $A$, we are done. Assume $S$ is countable now). Define a map $\phi$ from $\mathbb N \to S$, by $\phi(i) = n_i$. This is a bijection between $S$ and $\mathbb N$. Using composition, we see a bijection between $A$ and $\mathbb N$. – Sarvesh Ravichandran Iyer Dec 10 '16 at 04:02
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    I think this depends on how "countable" was defined. I that the being a one-to-one function was the definition so there is nothing to prove. (THis would be like saying "prove if n is divisible by two then it is even"-- you can't prove that as that is the definition.) So I am assuming your class uses a different definition. I can't tell you how to prove it without knowing what the definition is. – fleablood Dec 11 '16 at 01:11
  • @fleablood My definition of countable is a set that is either finite or denumerable. – Sam Dec 11 '16 at 01:19
  • Okay, but what does denumerable mean exactly? In my book it means in 1-1 corespondence to N. So... I don't say that anything needs to be proven. – fleablood Dec 11 '16 at 02:38
  • @fleablood oh, for my book I have denumerable means the set is equivalent to the natural numbers (N) – Sam Dec 11 '16 at 02:43
  • Okay.... then what does "equivalent" mean if it doesn't mean there is a one-to-one function between the two two sets that we are describing as "equivalent". Maybe, I should ask, does you class allow proofs of the form "x is given to have properties y. By definition z means property y. Therefore x is z. QED"? To my mind, that isn't a proof. It's a statement of knowing a definition. So, "by definition if the is a 1-1 function A to N, A is denumerable, and therefore by definition countible". If that's an acceptable proof.... go for it. – fleablood Dec 11 '16 at 02:52
  • @fleablood So it can be as simple as that? Honestly I think this is why I was confused. I thought too hard about this. But wait should there not be a bijection to prove denumerable? Is only one to one enough? – Sam Dec 11 '16 at 02:59
  • Well, no. I don't think it can be as simple as that. I'm as confused as you are because I don't think it is a proof if you state the definition. In such a case I don't think there is anything to prove. Basically we are simply told A is countable because having a 1-1 function to N is what countable means. So, I'm confused. – fleablood Dec 11 '16 at 03:07
  • Suppose you were asked if n is an integer so there is an integer k, so that n=2k, prove n is even". Do you think that is a fair question? I don't. So I don't think this question is fair either. – fleablood Dec 11 '16 at 03:14
  • @fleablood Yeah I am not too sure why this question is worded so poorly because to me it just doesn't really make sense. But thank you for explaining it a bit more. – Sam Dec 11 '16 at 14:48

3 Answers3

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If $f\colon A\rightarrow \mathbb{N}$ is one-to-one, then $f\colon A\rightarrow f(A)$ is bijective. Now, since $f(A)\subseteq \mathbb{N}$, then by this result $f(A)$ is countable, which gives us that $A$ is countable too.

Community
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Xam
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  • how are you getting that f:A--> f(A)? and that this is bijective? – Sam Dec 10 '16 at 04:33
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    $f\colon A\rightarrow f(A)$ is surjective because $f(A)={f(a): a\in A}$ and by the hypothesis $f:f\colon A\rightarrow \mathbb{N}$ is injective, so since $f(A)\subseteq \mathbb{N}$, then $f\colon A\rightarrow f(A)$ is also injective, hence it's bijective. – Xam Dec 10 '16 at 04:41
  • can I use the cardinality of A < or equal to N? – Sam Dec 11 '16 at 01:11
  • @Sam yeah sure. – Xam Dec 11 '16 at 02:22
  • can I use Rng (f) instead of f(A)? – Sam Dec 13 '16 at 02:37
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After much thought, I've decided this isn't entirely a triviality. If we assume the following definitions.

1) countable means either finite or denumerable. Finite means there is a bijection from some {1,....., n} to the set. Denumerable means there is a bijection from $\mathbb N$. (Frankly I never use the term "denumerable" and use "countably infinite" instead. Furthermore I assume in context that "countable" should be assumed infinite if not explicitly stated to be finite.)

2) 1-1 means injective but not surjective. I.e. for every $x \in A$ there is exactly one and only one $z \in f(A)$ so that $f(x) = z$. (Frankly, I never use 1-1 to mean injective and I always mean it to mean bijective. $f:A \rightarrow f(A)$ is always surjective and we can always for $f(A) \subsetneq X$ to make $f:A \rightarrow X$ not surjective if we wanted to so to say something misleading like $f:\mathbb R \rightarrow \mathbb R: f(x) = e^x$ is one to one because it is injective [but not surjective] is pointless and ... unsporting.)

So if we interpret the statement to be:

Prove: if an injection $f:A \rightarrow \mathbb N$ exists, then $A$ is countable.

That's not quite so trivial after all.

fleablood
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  • I was wondering if a proof by contradiction might be best in this case so I could use A is uncountable to work the proof out. – Sam Dec 11 '16 at 23:25
  • The best way is the obvious way of counting. Let C = f^-1(A) be the preimage of A. THen C is a subset of N. Argue that we can order any subset of N as {n_1, n_2,..... }. then let g(i) = f(n_i). That's a bijection from the indexes of C to A. Either the indexes are denumerable or finite. You have to argue some obvious things about sets of natural numbers--- that they can be ordered, that they are finite or countable, etc. – fleablood Dec 12 '16 at 00:29
  • It comes down to finding the first i->x of the preimage of A, relabeling it as 1->x, then finding the second j->y and relabeling it as 2->y, and so on. Formalizing it can be tedious. But it's mostly just definitions. – fleablood Dec 12 '16 at 00:32
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Let A be a non-empty set. Then the following are equivalent. (a) A is countable. (b) There exists a surjection $f :\mathbb N → A $. (c) There exists an injection $g : A → \mathbb N$. Proof. $(a) \Rightarrow (b)$ If $A$ is countably infinite, then there exists a bijection $f :\mathbb N → A $ and then (b) follows. If $A$ is finite, then there is bijection $h : ${$1,... ,n$} $→ A$ for some $n$. Then the function $f :\mathbb N → A $ defined by $$f(i) = \left\{ \begin{array}{ll} h(i) 1 \leq i \leq n\\ h(n) i > n. \end{array} \right.$$ is a surjection.

$(b)\Rightarrow (c)$. Assume that $f : \mathbb N → A $ is a surjection. We claim that there is an injection $g: A →\mathbb N$. To define $g$ note that if $a \in A$, then $f^{-1}$({a})$6=∅$. Hence we set $g(a) = \min f^{-1}$({a}). Now note that if $a \neq a'$then the sets $f^{-1}$({a})$\cap f^{-1}$({a′})= $∅$ which implies $\min^{−1}$({a})$\neq \min^{−1}$({a′}). Hence $g(a)\neq g(a')$ and $g : A → \mathbb N $ is an injective.

$(c)\Rightarrow (a)$. Assume that $g : A → \mathbb N$ is an injection. We want to show that $A$ is countable. Since $g : A → g(A)$ is a bijection and $g(A) \subset N$, and as we know that any subset of a countable set is countable, it implies that $A$ is countable.