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I know that null set is finite. But can you point me to a convincing proof?

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Do you mean the empty set, or a set of measure zero? Obviously e.g. $$\emptyset\subset \{1\}\implies |\emptyset|\le|\{1\}|=1,$$ which is obviously finite. And I'm doubtful this is really (number-theory). –  anon Nov 10 '11 at 4:51
Definition of finite? I would say a set is finite if there is a bijection between it and some $n$ where $n = \{0,\dots,n-1\}$ (these are formal ordinals defined inductively as $0=\emptyset$ and $m+1 = m \cup \{ m \}$). In this case, $0=\emptyset$ is finite by definition. –  Bill Cook Nov 10 '11 at 5:02
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3 Answers

up vote 5 down vote accepted

Take as the definition of "finite set" that $S$ is finite iff $S$ is in bijection with a set of the form $$\{k\in\mathbb{N}\mid k<n\}$$ where $n\in\mathbb{N}$. This is the definition used, e.g., on the nLab; the standard definition that $S$ is finite iff $S$ is in bijection with a set of the form $$\{1,\ldots,n\}$$ where $n\in\mathbb{N}$ (used, e.g., on Wikipedia) only necessitates defining the empty set to be finite, which is probably not satisfactory to you.

Now note that $\varnothing$ is in bijection with the subset $\varnothing=\{k\in\mathbb{N}\mid k<1\}$ (note that I am taking $\mathbb{N}$ to not include $0$). Hence, $\varnothing$ is finite. (If you prefer $0\in\mathbb{N}$, which there are many good reasons for, as Asaf points out below, and which many people do use, then of course we have $\varnothing=\{k\in\mathbb{N}\mid k<0\}$ instead.)

Alternatively, you could define an "infinite set" to be a set $S$ for which there exists a proper subset $T\subsetneq S$ which is in bijection with $S$; then define a finite set to be one which is not infinite. (As Asaf warns below, this is only an equivalent definition if one accepts the axiom of choice.)

Under this definition, because $\varnothing$ has no proper subsets, it cannot be infinite; hence it is finite.

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Zev, this is exactly the reason zero is treated as a natural number in set theory. Your definition of finite does not coincide with "for some $n\in\mathbb N$ the set has exactly $n$ elements". –  Asaf Karagila Nov 10 '11 at 6:01
Zev, your definition is fine but with $0\notin\mathbb N$ it does not agree with the one I said, which at times is easier to understand or work with. –  Asaf Karagila Nov 10 '11 at 6:12
@Asaf: I've edited to reflect your point. I agree that having that definition be true would be preferable. Another good reason is that it'd be nice if $\mathbb{N}$ were the free monoid on one element. But I am still debating whether to "convert" :) –  Zev Chonoles Nov 10 '11 at 6:13
Another nitpick, your definition of infinite is known as Dedekind-infinite. It is known that it requires the axiom of choice (slightly less than countable choice) to ensure that this definition is equivalent to "$X$ is infinite if it is not finite"... –  Asaf Karagila Nov 10 '11 at 6:32
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The definition of an infinite set is such:

$A$ is infinite if and only if for every $n\in\mathbb N$ there exists a function from $\{0,\ldots,n-1\}$ to $A$ which is 1-1 (i.e. injective).

This agrees with the definition $A$ is infinite if and only if it is not finite; whereas finite is defined to have $n$ many elements for some $n\in\mathbb N$.

From the above definition we can easily see why $\varnothing$ cannot be infinite, there are no injective functions from $\{0\}$ into $\varnothing$. In fact, there are no functions at all from a nonempty set into the empty set.

However, one could argue that sets are either empty, finite, or infinite. Of course this all depends on the definition, and to a certain degree matters less - as long as you cannot deduce that $\varnothing$ is infinite. It is important that definitions that have become accepted in the mathematical world are used by most people, and if you deviate from them then you have to give your definition as well, as to avoid confusion.

In the case that you want to take the definition of the empty set as non-finite, but rather empty, then you have to reformulate most of the theorems regarding finality, for example:

Theorem: Every subset of a finite set is finite.

Now this would have to read: "Every subset of a finite set is finite or empty." because the empty set is a subset of every other set. If you are willing to reformulate, give explicit definitions and so on, then it is okay to use "nonstandard" terminology and notation. If not, then it should be apparent that the conventional norms should be fine, and under those the empty set is indeed finite.

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Since I generally take ZFC as my framework, I prefer first to define the finite ordinals and then to define finite sets as those that admit a bijection with a finite ordinal. The finite ordinals are the elements of $\omega$, which is defined using the axiom of infinity and the comprehension schema. There are a couple of ways to carry out the details, but in the end $\omega$ turns out to be the intersection of all sets $x$ such that (a) $\varnothing \in x$ and (b) $y\cup\{y\}\in x$ whenever $y\in x$, i.e., the smallest set containing $\varnothing$ and closed under the successor operation $y\mapsto y\cup\{y\}$. In particular, $\varnothing \in \omega$, so by definition $\varnothing$ is finite.

In this approach we go on to define $0$ to be $\varnothing$, $1$ to be $\varnothing\cup\{\varnothing\}=0\cup\{0\}$, $2$ to be $1\cup\{1\} =$ $0\cup\{0\}\cup\{1\}=\{0,1\}$, and so on, and each ordinal is the set of smaller ordinals. Thus, a bijection between a set $A$ and a finite ordinal $n$ is in fact a bijection between $A$ and $\{k:k<n\}$, so this approach actually leads to the definition in Zev’s answer in the version in which $\mathbb{N}=\omega$, i.e., includes $0$.

There are other ways to define finite set, but all of the ones that I’ve seen automatically make the empty set finite. For instance, a definition due to Kuratowski can be expressed as follows. Let $S$ be any set. Let $$\mathfrak{F}(S) = \{\mathscr{A}\subseteq\wp(S):\varnothing\in \mathscr{A}\land\forall X\in\wp(S)\;\forall x\in S\big(X\in\mathscr{A}\to X\cup\{x\}\in\mathscr{A}\big)\},$$ and let $\mathscr{F}(S)=\bigcap\mathfrak{F}(S)$; then $S$ is finite iff $S\in\mathscr{F}(S)$. Applying this definition to the empty set, we have $\wp(\varnothing)=\{\varnothing\}$, whose subsets are $\varnothing$ and $\{\varnothing\}$, of which only $\{\varnothing\}$ belongs to $\mathfrak{F}(\varnothing)$. Thus, $\mathfrak{F}(\varnothing)=\{\{\varnothing\}\}$, $\mathscr{F}(\varnothing)=\bigcap\{\{\varnothing\}\}=\{\varnothing\}$, and hence $\varnothing\in\mathscr{F}(\varnothing)$, i.e., $\varnothing$ is finite.

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