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
 A: *

*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.

*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."
A: "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.
A: 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 $y=x$.
A: (This late answer expands on lhf's answer.)
Here is the analogy which I see.
"If and only if" is the natural language description of two implications in the opposite directions:
\begin{eqnarray}
\tag{if and only if}
P \;\color{red}\equiv\; Q \\
\tag{if}
P \;\Leftarrow\; Q \\
\tag{only if}
P \;\Rightarrow\; Q \\
\end{eqnarray}
Now, when talking about existence and uniqueness of mathematical objects, then the simplest ways to express these are as follows:
\begin{eqnarray}
\tag{one and only one}
\langle \exists! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \color{red}\equiv & x = y \rangle \rangle \\
\tag{one}
\langle \exists\phantom! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \Leftarrow & x = y \rangle \rangle \\
\tag{only one}
\langle \phantom\exists! x :: P(x) \rangle & \;\equiv\; & \langle \exists y :: \langle \forall x :: P(x) & \Rightarrow & x = y \rangle \rangle \\
\end{eqnarray}
And the analogy is visually clear.
(Here $\langle ! x :: P(x) \rangle$ is my notation for "there is at most one $x$ such that $P(x)$".  Apologies for the duplication from another question of mine.)
