Unique Existence - Existence and Definitions

Mathematical Writing - Vivaldi Franco 2014

Unique Existence
Existence and Definitions

Unique existence is a stronger version of existence:

There is exactly one element of $$X$$ having property $${\fancyscript{P}}$$.

where, as usual, $$X$$ is a set and $${\fancyscript{P}}$$ is a predicate over $$X$$. For example:

The equation $$f(x)=0$$ has exactly one solution.

The sets $$A$$ and $$B$$ have only one point in common.

The function $$g$$ has precisely one stationary point.

The matrix $$M$$ has a unique real eigenvalue greater than 1.

The expressions ’exactly’, ’only’, ’precisely’ differentiate unique existence from existence.

Unique existence is a hidden conjunction; the conjuncts are

EXISTENCE: There is one element of $$X$$ with property $${\fancyscript{P}}$$.

UNIQUENESS: If $$x,y\in X$$ have property $${\fancyscript{P}}$$, then $$x = y$$.

The corresponding symbolic expression is overloaded:

$$\begin{aligned} (\exists x\in X,{\fancyscript{P}}(x))\,\wedge \, (\forall x,y\in X,({\fancyscript{P}}(x)\wedge {\fancyscript{P}}(y))\,\Rightarrow \, (x=y)) \end{aligned}$$

(9.1)

and to simplify it we introduce a new quantifier $$\exists !$$ denoting unique existence:

$$ \exists !\, x\in X,{\fancyscript{P}}(x). $$

For example, given a function $$f{:}\,X\rightarrow Y$$, the concise expression

$$ \forall y\in Y,\exists !\, x\in X,f(x)=y $$

states that $$f$$ is bijective.

In Chap. 8 we proved (twice) that any integer $$n$$ greater than 1 is a product of primes. For each $$n$$, this statement asserts the existence of an unspecified number of primes. The fundamental theorem of arithmetic upgrades existence to unique existence .

Theorem

Every natural number greater than 1 may be written as a product of prime numbers. This representation is unique, apart from re-arrangements of the factors.

The expression ’apart from re-arrangements of the factors’ gives the impression of lack of uniqueness. This is not the case: the theorem states that there exists a unique multiset of prime numbers (see Sect. 2.1) with the property that the product of its elements is the given natural number. Alternatively, there is a unique set of pairwise co-prime prime-powers with the same property.

In a proof of unique existence the conjuncts are normally proved separately, in either order.

Theorem

The identity element of a group is unique.

The group axioms were given in Sect. 7.1. The existence of an identity element does not require a proof, being an axiom (axiom G3). We only have to prove uniqueness .

PROOF Let us assume that a group has two identity elements, $$\diamondsuit _1$$ and $$\diamondsuit _2$$. Applying the axiom G3 to each identity element, we obtain

$$ \diamondsuit _1 \odot \diamondsuit _2= \diamondsuit _1 \qquad \diamondsuit _1 \odot \diamondsuit _2= \diamondsuit _2 $$

and hence $$\diamondsuit _1=\diamondsuit _2$$. $$\square $$

Theorem

For every set $$X$$ and subset $$Y$$, there is a unique set $$Z$$ such that $$Y \cup Z = X$$ and $$Y\cap Z = \emptyset $$.

PROOF Let $$Z=X{\backslash }Y$$. By construction, $$Y\cap Z = \emptyset $$. Since $$Y\cup Z$$ is a subset of $$X$$, and every element of $$X$$ belongs to either $$Y$$ or $$Z$$, we have $$Y \cup Z = X$$. So a set $$Z$$ with the required properties exists.

To prove uniqueness, assume that $$Z$$ is a set with the stated properties. Then, since $$Y \cup Z = X$$, $$Z$$ must be a subset of $$X$$, and it must contain all elements of $$X$$ that aren’t in $$Y$$. So $$Z$$ must contain $$X{\backslash }Y$$.

Since $$Y \cap Z = \emptyset $$, the set $$Z$$ can’t contain any elements of $$X$$ that are in $$Y$$. So $$Z$$ must be $$X{\backslash }Y$$. $$\square $$

The existence part of this proof is constructive. For this reason, the proof of uniqueness does not establish the second conjunct in (9.1) explicitly. Rather than considering two objects with the stated properties and showing that they are the same, we consider a single object and show that it is the same as the one defined in the existence part.

Theorem (Fermat). The only natural numbers $$x$$ for which $$1+x+x^2+x^3$$ is a square are $$x=1$$ and $$x=7$$.

Proving existence is immediate: $$1+1+1^2+1^3=2^2$$, $$1+7+7^2+7^3=20^2$$. Uniqueness is much harder, see [12, pp. 38, 382].