So I am not exactly sure how to prove this out rigorously. Intuitively it makes sense but I feel like I am misinterpreting it because it seems rather trivial to me. Any help is much appreciated.
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The (finite) cardinal of the proper subset is strictly lesser than the (finite) cardinal of the finite set. – Lucian Jan 30 '14 at 06:23
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And your definition of infinite set is...? – Martín-Blas Pérez Pinilla Jan 30 '14 at 23:34
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I think the usual convention is to call a set finite if there is a bijection between the set and some natural number. – hot_queen Jan 30 '14 at 23:36
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If you study axiomatic approach to set theory, then it would be helpful if you post axioms or whatever you have studied so far.
A lot of things about finite sets are easier to prove by induction, because this is at least one way to define finite sets. Here is a sketch of a proof based on this approach:
1) A finite set is by definition the one that is in one-to-one correspondence with a set of natural numbers $\{1,\dots,n\}$ for some $n$, where natural numbers are defined by induction. $n$ is called the cardinality of the set. So, the question becomes whether such a set can be in one-to-one correspondence with its own proper subset.
2) If $n=1$ or $n=2$ you can check (prove) manually that this is not the case.
3) Now, suppose that this cannot be done for all sets with cardinality up to $n\ge 2$. Consider a set with cardinality $n+1$: $A=\{1,\dots,n+1\}$. And let $f$ be a bijection from $A$ onto its proper subset $B$. Since $B$ is a proper subset of $A$, there is some $k$ such that $k\not\in B$. Let $f(k)=m$, then $m\neq k$, and $f$ restricted to $A'=A-\{k\}$ is a bijection such that neither $m$ nor $k$ is in the image, so that it is a bijection from $A'$ onto its proper subset. Now, it remains to notice that $A'$ has cardinality $n$.
Vadim
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Let $A$ be a finite set, and let $B$ be a proper subset of $A$.
The existence of a one-to-one correspondence between the elements of $A$ and $B$ implies that the cardinality of $A$ equals the cardinality of $B$:
$$|A| = |B|$$
However, since $B$ is a proper subset of $A$, it must have a smaller cardinality. This contradicts $|A| = |B|$, so there can't exist a one-to-one correspondence.
Zubin Mukerjee
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All you've done is mask your assumption that $|A|<|B|$. This is not helpful (though if you turned it in it may squeak by a tired homework grader). – Jan 30 '14 at 06:35
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Hint
If $A$ is a finite set with cardinal $n\in\mathbb N$ then it's on bijection with the set $\mathbb N_n=\{1,\ldots,n\}$.
To see the desired result we have to prove that $$ \text{there's a bijection }\; f\colon \mathbb N_p\rightarrow \mathbb N_n\iff n=p$$ by proving
- $f$ is an injection iff $p\le n$
- $f$ is a surjection iff $p\ge n$.
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1Wait, $p \leq n$ doesn't tell you $f$ is injective, $f$ could be constant. – Dylan Yott Jan 30 '14 at 06:37
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@DylanYott I mean by the first point: $$\text{there's an injection }; f\colon \mathbb N_p\rightarrow \mathbb N_n\iff p\le n$$ – Jan 30 '14 at 15:44
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This is due to the pigeonhole principle, but I'm not quite sure how one would prove it. Perhaps by induction?
Henry Swanson
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It would take a slight change in the original statement for the pigeonhole principle to be directly applicable. See my first answer here. – Dan Christensen Jan 30 '14 at 23:17
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Suppose that $A\subsetneq B$ and that $f:B\to A$ is a bijection. Take $x_{0}\in B\setminus A$, and define a sequence recursively by $x_{k+1}=f(x_{k})$. It is easy to prove that this sequence never repeats, and so $B$ must contain a copy of $\mathbb{N}$ and thus be infinite.
Unwisdom
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Define a sequence sets recursively: $A_0=B $ and $ A_{k+1}=f [A_{k}] $. Then the set ${A_{k}\setminus A_{k+1}\vert k\in\mathbb {N}}\cup\left{\bigcap_{k=0}^{\infty} A_k\right} $ is a partition of $ B $, with $ x_k\in A_k $. – Unwisdom Jan 30 '14 at 23:34
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