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As a beginner I am not quite settled on why unique existence proofs have to show two things.

In general, to show there exists a unique $x$ that conforms to a property $P(x)$ we show:

  1. $x$ conforms to property $P(x)$

  2. $P(x)$ entails (implies?) $x$.

Is this a good way to think about it?

If yes, then I'm happy.

Example:

Show there exists a unique integer $x$ such that $x + 3 = 5$.

Let's suppose $x=2$ is the unique integer.

The proof steps are then:

  1. Show $x=2$ conforms to the property. Indeed $(2) + 3 = 5$ is a true proposition.

  2. Show $P(x)$ implies $x=2$. We use algebra $x+3 = 5$ is equivalent to $x +3-3 = 5-3$, which is equivalent to $x=2$.

The reason I am unsure about unique existence proofs is that at school we only did step (2) above. We derived the solution, and the algebra told us how many solutions there were, eg quadratic equations had 2 solutions sometimes.

Note I asked a related question last year, but this question is asking about a generic way to think about all unique existence proofs.

Penelope
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    To prove uniqueness assume there exists some $2'$ so that $2'+3=5$ and then show $2'=2$. It's more like the second approach you used. – CyclotomicField Feb 12 '25 at 23:59
  • It's certainly true that when "evaluating" expressions such as $\sqrt2^{\sqrt2^{\sqrt2^{\cdot^{\cdot^{\cdot}}}}}$—which we might interpret as something like $x_0 = 1, x_k = \sqrt2^{x_{k-1}}, \lim_{k\to\infty} x_k = ?$—we feel compelled to show not only what the limit must equal if it exists (which is like your #2) but also that it exists (which is like your #1). After all, if the value doesn't exist, it can't equal anything. So I tend to think that both sides are necessary, yes. – Brian Tung Feb 13 '25 at 00:03
  • I think the answers to your related question do in fact address your quest for a "generic way" to think about "there is a unique ..." assertions. Please [edit] this question to show us in more detail why the other answers don't suffice. An example would help (even though you are asking about a generic situation). – Ethan Bolker Feb 13 '25 at 01:23

1 Answers1

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The statement "There exists a unique $x$ such that $p(x)$" has two parts:

  • There is at least one $x$ such that $p(x)$. (Existence)
  • There is at most one $x$ such that $p(x)$ (Uniqueness)

Together, they establish that there is exactly one x such that p(x).

Uniqueness tells you that there is at most one solution; if you have proven uniqueness but you haven't yet established whether there is a solution, you know that there is either exactly one solution, or no solution at all.

I think this is an awkwardness in the mathematician's choice of terminology—in everyday language, saying something is unique strongly suggests that it exists. In mathematics, it is instead shorthand for saying that there is either zero or one object with this property (i.e. that all objects with this property are equal, perhaps vacuously so.)

Here is an example of a uniqueness proof where the solution in fact doesn't exist:

  • The maximal integer $n\in \{1,2,3,\ldots\}$, if it exists, is unique. Proof: For all $n > 1$, we know that $n^2 > n$. Therefore $n$ cannot be the largest integer ($n^2$ is larger). Therefore, the largest integer—if there is any—must be equal to one.
user326210
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  • yes - this is indeed the cause of confusion "saying something is unique strongly suggests that it exists" – Penelope Feb 13 '25 at 11:27