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I'm trying to solve a problem from Kenneth O. Rosen Discrete Mathematics and its Applications:

Show that set S is infinite if and only if there is a proper subset A of S such that there is one-to-one correspondence between A and S.


Proof:

Let $S$ be finite so it's cardinality is equal to $n$, where $n \in \mathbb{N} $. Then from the fact that $A$ is a proper subset of $S$ it follows that $|A| < n$. But since $S$ and $A$ have one-to-one correspondence $|A|=|S|=n$. Contradiction, therefore $S$ is infinite.


But I am not really sure if I've proven it correctly. To me it's paradoxical that a set that is smaller than it's superset can have bijection with it and I don't understand how that is possible. I did read some similar questions:

But still could not understand. The book does not give notion of infinite sets at least not now.

Did I prove this problem correctly? How is it possible that a set that is smaller has all elements of its superset?

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  • $\begingroup$ "To me it's paradoxical that a set that is smaller than it's superset can have bijection with it" One of the first and simplest examples is understanding that $\{1,2,3,4,5,\dots\}$ is in bijection with $\{2,4,6,8,10,\dots\}$ with the very clear bijection $f~:~\Bbb N\to 2\Bbb N$ given by $f(n)=2n$. That is to say... there is a "first" even number, a "second" even number, a "third" even number and so on... $\endgroup$
    – JMoravitz
    Jul 30, 2023 at 15:03
  • $\begingroup$ What if n = 0, can you still find such an A? I believe you are confounding what assumptions you are working under. Either that, or I misunderstood your proof. For a "if and only if" proof you need to prove both directions. Direction 1: Assume S is infinite, show that for some strict subset A, there is a bijection from A to S. Direction 2: Assume we have a bijection from Such an A to S, prove S is infinite. $\endgroup$ Jul 30, 2023 at 15:19
  • $\begingroup$ @MichaelCarey my bad, I should've specified that n is some natural number, because for n = 0, the whole question and proof does not make sense. $\endgroup$
    – Armagidon
    Jul 30, 2023 at 15:23
  • $\begingroup$ @DumperDGarb since I don't yet have notion of infinite sets, not to mention any axioms/theorems I am just assuming my statements to be true for finite sets, and since things don't work out when these are finite sets, $S$ should be infinite. $\endgroup$
    – Armagidon
    Jul 30, 2023 at 15:28
  • $\begingroup$ @DumperDGarb thank's. That's why I am asking. The book apparently expects me to know this, or doesn't but does not give any information about what kind of reasoning should I use. $\endgroup$
    – Armagidon
    Jul 30, 2023 at 15:39

3 Answers 3

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Note the use of "if and only if" in the question. This is logically equivalent to proving the following:

"If $S$ is infinite, then there is a proper subset $A$ of $S$ such that there is a one-to-one correspondence between $A$ and $S$, and if there is a proper subset $A$ of $S$ such that there is a one-to-one correspondence between $A$ and $S$, then $S$ is infinite."

First prove one conditional (by whatever means you like), and then prove the other conditional (by whatever means you like). Your proof is correct, but it's only half complete.

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No, this is not a complete proof. 'If and only if'-statements are always two statements disguised as one. You need to show two directions:

  • First you need to show that if $S$ is infinite, then a proper subset $A$ and a bijection from $S$ to $A$ exist.

  • Then you show that if $S$ has a proper subset $A$ and a bijection $S$ to $A$, then $S$ if infinite.

Both of these statements can be proven directly or by contradiction. You have already proven the second statement.

Here you have a complete proof. The first part is your proof (slightly rephrased).

Let $S$ be a set, let $A$ be a proper subset of $S$ and $f: A \rightarrow S$ a bijection. Assume that $S$ is finite. $|A| < |S|$ holds true because $A$ is a proper subset of $S$. But since there is a bijection between $S$ and $A$, $|A|=|S|$ also holds true. So there is a contradiction, therefore $S$ must be infinite.

Now let $S$ be an infinite set. Let $T$ be an countably infinite subset of $S$. We can w.l.o.g. assume that $T = \mathbb{N}$, because for every countably infinite set there is a bijection to $\mathbb{N}$. Now let $T' = \mathbb{N} \setminus \{ 1 \}$ and $g: T' \rightarrow \mathbb{N}, n \mapsto n - 1$. $g$ is a bijection because every element of $\mathbb{N}$ has a corresponding element in $T'$ and vice versa. Now let $A = S \setminus \{ 1 \}$ and $f: A \rightarrow S, x \mapsto f(x)$, where $f(x) = g(x) = x - 1$ if $x \in T'$, otherwise $f(x) = x$. $f$ is a bijection because on $T' \subset S$ it acts like $g$ which is a bijection, on $S \setminus T'$ it is the identity and $f(A) = S$.

Did I prove this problem correctly? How is it possible that a set that is smaller has all elements of its superset?

No, this is not possible. But a one-to-one correspondence does not mean that every element in one set is also in the other. It means that you can associate every element from one set with exactly one unique element from the other set, without leaving any elements unassociated. In my proof I constructed a one-to-one correspondence between $\mathbb{N} \setminus \{ 1 \}$ and $\mathbb{N}$ via the function $n \mapsto n-1$. $\mathbb{N} \setminus \{ 1 \}$ is clearly a proper subset of $\mathbb{N}$, but still you can construct a bijection. This is a unique property of infinite sets, that does not apply to finite sets.

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  • $\begingroup$ I don't yet know about enumerable sets, so maybe what you did with them is fine, but why did you assume that $S$ contains numbers? Also I could not understand why did we need $T$? As far as I understood we could've proved this without $T$. Can we prove this without enumerable sets? The book only introduces to regular sets from naive set theory. $\endgroup$
    – Armagidon
    Jul 30, 2023 at 16:03
  • $\begingroup$ @Armagidon I guess countable is the more common term. $T$ is a countably infinite subset of $S$. (I forgot the infinite in the answer, will edit that in.) Countable infinity is the smallest kind of infinities, that's why that subset must exist for every infinite set $S$. Now for every countably infinite set there is a bijection to $\mathbb{N}$. That is the defining property of countably infinite sets. That's why I assumed without loss of generality that $T$ is $\mathbb{N}$, and then later in the proof just referred to it as $\mathbb{N}$. $\endgroup$
    – chrysante
    Jul 30, 2023 at 16:14
  • $\begingroup$ So $S$ does not necessarily contain numbers, but there always is a bijection $h$ from $\mathbb{N}$ to a subset of $S$, and therefore $S$ contains the elements $h(n)$ where $n \in \mathbb{N}$. I just treated $n$ and $h(n)$ as the same thing. $\endgroup$
    – chrysante
    Jul 30, 2023 at 16:19
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Consider the discrete case, the set of all natural numbers $N$. This set includes all integers greater than 0. Here are the first ten elements of the set: $$ N=\{1,2,3,4,5,6,7,8,9,10...\} $$ We can then remove all odd numbers to get a new set $P$: $$ P=\{2,4,6,8,10...\} $$ We only have $5$ elements now, but we can continue the pattern to get $10$. We can do this only because $|N|$ is a countable infinity. $$ P=\{2,4,6,8,10,12,14,16,18,20...\} $$ However, this result is the same as the result if all elements in $N$ were scaled by a factor of $2$. See: $$ 2N=\{2,4,6,8,10,12,14,16,18,20...\} $$ Thus, $P=2N$. However, since scaling all the elements of a set by a single (nonzero) factor does not change the cardinality, we must conclude that since $|N|=|2N|$ and $|P|=|2N|$, $|P|=|N|$.

Thus, your proof is correct. Bear in mind that the line of reasoning I presented is designed to be an intuitive explanation, not a rigorous one.

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  • $\begingroup$ So if I remove one element from a set with cardinality of countable infinity they would still be equal? Is it because $\infty-1 = \infty$? $\endgroup$
    – Armagidon
    Jul 30, 2023 at 15:43
  • $\begingroup$ @Armagidon Its not that simple. Infinities are a difficult concept, and require more rigorous proofs that simply subtracting cardinalities from one another. Usually, the best way to prove that a bijection exists between two sets is to find a function with a $1$ to $1$ mapping between those two sets. For instance, in reference to your comment, if the $1$ was removed from the set of natural numbers $N$, forming set $P$, there would still exist a bijection because the function $f(n)=n+1$ would be able to $1$ to $1$ map all elements of $N$ to $P$ with no duplicates or skipped numbers. $\endgroup$ Jul 30, 2023 at 15:54

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