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For example, an injective function can be defined as

$\forall y \in Y$ there exists at most one $x \, : \, \forall f(x) = y$

How could I represent the bold part of the above expression? For "there exists at least one" we have $\exists$, for "there exists a unique" we have $\exists!$. Is there a symbolic notation for "...at most one"? The classic definition of injective, i.e., $\forall \, a,b\in X, \, f(a) = f(b) \rightarrow a = b$ seems to side-step this, though.

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  • $\begingroup$ I'd be very surprised if here was a symbol for the concept "at most one", I've never seen it. $\endgroup$
    – Git Gud
    Jan 21, 2017 at 17:40
  • $\begingroup$ Maybe expressible in a non-conventional logic system such as modal logic. To have a short idea of a certain number of logic systems, have a look at (en.wikipedia.org/wiki/Intuitionist_logic) $\endgroup$
    – Jean Marie
    Jan 21, 2017 at 22:20

3 Answers 3

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If you will be using it repeatedly, you could invent a notation... $$ \forall y \in Y\;\exists^{\le 1} x $$ Of course you will have to define it, say like this $$ \exists^{\le 1} x\;\;\Phi(x)\qquad\text{means}\qquad \forall x_1 \forall x_2 \big(\Phi(x_1) \wedge \Phi(x_2) \longrightarrow x_1=x_2\big) $$

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  • $\begingroup$ This is a very good idea...! A subscript seems to be clear and easy to understand. $\endgroup$
    – Masacroso
    Jan 21, 2017 at 17:44
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I have never heard of such a symbol, but if you really want to you can write $\exists !\, x \lor \nexists\, x, \,\dots $

However, this is not easily readable and I would not advise its use at all.

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For example, an injective function can be defined as

$\forall y {\in} Y$ there exists at most one $x : \, f(x) = y$

How could I represent the bold part of the above expression? For "there exists at least one" we have $\exists$, for "there exists a unique" we have $\exists!$. Is there a symbolic notation for "...at most one"? The classic definition of injective, i.e., $\forall \, > {a,b}\in X, \, f(a) = f(b) \rightarrow a = b$ seems to side-step this, though.

If you accept a string of symbols in lieu, then how about:

$$\forall y{\in}Y \; \exists p\;\forall x{\in}X\:\Big(f(x)=y \implies p=x\Big). $$

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  • $\begingroup$ This is incorrect. You are defining unique existence. $\endgroup$ Nov 28, 2022 at 2:29
  • $\begingroup$ @SimonEatwell Nope: "exactly one" would instead be $\exists p\,\forall x\:\big(Px\leftrightarrow p{=}x\big).$ $\endgroup$
    – ryang
    Jan 28, 2023 at 0:25
  • $\begingroup$ You are right. How stupid of me! $\endgroup$ Jan 29, 2023 at 9:01
  • $\begingroup$ @SimonEatwell You weren't being stupid. I've just edited the answer for clarity. $\endgroup$
    – ryang
    Jan 29, 2023 at 10:02

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