# Formal way to state that a condition is true for a certain number of elements of a set.

This may be very simple, but I have the following problem: I have a certain logical condition, let us call it $$\theta(i)$$, with $$i \in X$$, which can be true or false for the argument $$i$$. Now I want to formally state that $$\theta(i)$$ cannot be true for more than one element of the set $$X$$ (being true for none of the elements of $$X$$ should also be included). Can anyone help me?

• $\left|\{i\in X\,|\,\theta(i)\}\right| \le 1$
– Matt
Commented Aug 1 at 13:40

I suggest a frame change. If you are writing for a human reader and not a software application that checks proofs written in some formal language, accurate words are easier to understand than logical propositions, and just as rigorous. So simply say

$$\theta(i)$$ is true for at most one element $$i \in X$$.

There are multiple ways to state this, but one that works well in constructive mathematics is to say that the subset of $$X$$ on the elements which satisfy $$\theta$$ is a mere proposition, in the sense that any two of its elements are equal:

$$\forall i, j \in X. \theta(i) \land \theta(j) \to i = j$$

or equivalently

$$\forall i, j \in \{x \in X \mid \theta(x)\}. i = j$$

• Just for interpretation: this means that $\theta(i)$ can only be true for two elements of $X$ if the two elements are equal, right? Commented Jul 31 at 12:31
• Yeah, or rather the meaning of that sentence is what I wrote :) Commented Jul 31 at 12:33
• @zinsinho: ... Or in other words, it's not two elements, it's one single element. Commented Aug 1 at 0:54

For simplicity, how about stating there is not a distinct $$i$$ and $$j$$ such that $$\theta$$ is true for both, i.e.: $$\not\exists i, j \in X, i \not = j : \theta(i) \land \theta(j)$$

In the general case if $$Y$$ is your cardinal you just want $$X \cap\theta$$ in bijection with $$Y$$, because bijection is 'is a certain number' for cardinals.

You can expand each bit to however formal you want it but I'd stop there, because expressing the existence of a bijection in terms of $$\in$$ and $$\mathcal{P}$$ is an exercise you only need to do once.

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