# Analogy between “One and only one” and “If and only if”

Is there a close analogy between "one and only one", used to mean exactly one (as in "there is one and only one object satisfying that condition" etc.), and "if and only if"? "If" and "Only if" can be used separately to denote sufficient and necessary conditions, but I'm not sure whether the constituent parts of "one and only one" can be considered in isolation.

I think it looks like you could interpret "there is one X satisfying condition Y" as "there is one or possibly more X satisfying condition Y", but I am not sure about the difference in interpretation between "there is only one X satisfying condition Y" and "there is one and only one X satisfying Y". Could "there is only one X satisfying condition Y" be interpreted as "there could be one X, but possibly none satisfying Y"? I would appreciate any ideas on this. Thanks

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Why would anyone close this perfectly reasonable question? –  Phira Nov 29 '12 at 18:23
@Phira Hi Phira, is the question still closed? What was the reason for this, as I could try and improve the question if possible? –  user50229 Nov 30 '12 at 13:06
It isn't closed, there are only two votes in favour of closing, so I commented to make other voters think twice. –  Phira Nov 30 '12 at 16:10

1. To show the existence of an X satisfying Y it is sufficient to prove "existence" of such an X: that there is "some" (one or more) X satisfying Y.

2. Then, to show that there is only one such X, you need to show that if K also satisfies Y, then K must equal X.

So "one and only one" requires establishing both (1) and (2): existence and uniqueness.

Put differently, "there is one and only one" can be read as the conjunction of:

(a) existence of "at least one" X such that X satisfies Y,

... and ...

(b) existence of at most one such X that satisfies Y.

If we let $P(x)$ denote the satisfaction of some property $P$ by $x$, then we can assert that there is one and only one $x$ such that $P(x)$ as follows:

$$\exists x[P(x) \land \forall y(P(y) \rightarrow y = x)]$$

For the difference between: (i) "only one X satisfies condition Y" and (ii) "there is one and only one X satisfying Y": One can argue that there might be some property that only one element could possibly satisfy, without necessarily asserting that therefore, such an element exists (ii).

Using "there is one and only one" or "there exists a unique" (or even "there exists exactly one") is less ambiguous than stating "only one", which can be taken to mean "at most one."

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I think I was asking along the lines of your second alternative in my question: Does this mean that if I say in isolation "there is only one such X" that this means "there is at most one such X" so technically there could be none? –  user50229 Nov 29 '12 at 17:05
If you say "there is only one", you are asserting existence ("there is...") and uniqueness (...only one). If you say "Only one X could be Y", then you are saying "at most, one X could be Y." –  amWhy Nov 29 '12 at 17:11
Since 'only one' alone means 'at most, one X could be Y', does this mean that it could only be separated as I suggested if we consider modal logic? –  user50229 Nov 30 '12 at 21:16
It can be separated without modal logic: only one corresponding to at most one, and existence (there is...) corresponding to at least one. –  amWhy Nov 30 '12 at 21:20
I like this wikipedia page which has a line where the two conditions are shown separated out $\exists x P(x) \wedge \forall y\, \forall z\,((P(y) \wedge P(z)) \to y = z)$. I think you could identify 'one' with the first clause and 'only one' with the second clause? Anyway I will accept your answer because of your detailed help, thanks –  user50229 Nov 30 '12 at 21:27

They are related because the statement "There is one and only one $x\in U$ such that $P(x)$" is equivalent to "There is $x\in U$ such that $P(y)$ if and only if $y=x$.

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"If and only if" is a two-way implication, i.e. the left implies the right and the right implies the left. "One and only one" is a statement regarding existence and uniqueness. There is not necessarily a two-way relation here. For instance, the solution to a quadratic can exist, but need not be unique.

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Uniqueness implies existence, so does this mean "only one" and "one and only one" are the same, and it is just a matter of emphasis? –  user50229 Nov 29 '12 at 16:58
@SJGreen Uniqueness does not imply existence... it would be very nice if it did. For example, the Collatz path from some arbitrary (very large) integer $n$ to 1 would be unique, but it is not known to exist for all $n$. –  Arkamis Nov 29 '12 at 17:04
In this sense, "one and only one" probably could be thought of as a two way implication, "if it exists, it is unique; if it is unique, it exists." I am not sure how common this usage is. –  Arkamis Nov 29 '12 at 17:06
Yes I see where my comment was wrong, an object could be unique if it existed, so it looks like proving an object is the "only one" is necessary but not sufficient for there being "exactly one"? I think this is where the analogy with "if" and "only if" comes in. –  user50229 Nov 29 '12 at 17:09