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On the Wikipedia page for the Axiom of Choice the following statement is given:

$(\forall x^\sigma)(\exists y^\tau)R(x,y)\rightarrow(\exists f^{\sigma \rightarrow \tau})(\forall x^\sigma)R(x, f(x))$

Most of it seems fairly straightforward, except for the meanings of the symbols that look like 180 degree rotated 'E' and 'A'

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    $\begingroup$ Interesting question. I don't really understand how this can be considered "fairy straightforward" if the 4 quantifiers aren't known or understood. $\endgroup$ Jan 11, 2020 at 13:20
  • $\begingroup$ See here & here. $\endgroup$
    – J.G.
    Jan 11, 2020 at 15:09
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    $\begingroup$ It's worth adding that their LaTeX codes are "\forall" and "\exists" respectively. $\endgroup$ Jan 12, 2020 at 17:54

2 Answers 2

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The symbols are quantifiers. They bind a new variable name to the symbolic logic statements. ∃ reads as there exists. ∀ reads for all so the first part of the statement would be read as:

forall x (of type 𝜎), there exists a y (of type 𝜏) such that ...

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    $\begingroup$ As someone who thought OP's quote looks like caveman writing, could you expand this answer to explain all the hieroglyphs? $\endgroup$ Jan 11, 2020 at 1:47
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    $\begingroup$ @user1717828 I'm pretty sure the OP couldn't understand anything too given that they did not know the most basic elements of the formula (which they call "equation" even though there is no equal sign anywhere...). $\endgroup$
    – Bakuriu
    Jan 11, 2020 at 8:38
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    $\begingroup$ @user1717828 I'll move to Roman letters: "Let $R$ be a relation of things in $X$ to things in $Y$. If it is the case that for all $x$ in $X$ there is a $y$ in $Y$ which is related to $x$, then there is a function $X \to Y$ which takes any $x$ to a thing which is related to $x$." A relation of things in $X$ to things in $Y$ is viewed as a predicate on $X \times Y$, which answers "True" to the input $(x, y)$ if and only if $x$ is related to $y$; hence the use of the function notation $R(x, f(x))$ which means "$x$ is related to $f(x)$". $\endgroup$ Jan 11, 2020 at 8:41
  • $\begingroup$ @Bakuriu indeed. I also had the same mistake. fixed it now. $\endgroup$
    – Apoorv
    Jan 11, 2020 at 9:10
  • $\begingroup$ @Bakuriu I figured out how to interpret most of it, I just don't have the background knowledge to know that 'equation' isn't the right term. $\endgroup$
    – 0x777C
    Jan 12, 2020 at 16:49
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$\forall$ reads as "for all", and $\exists$ reads as "there exists". So, in english we have

$$\text{"if }\underbrace{\text{for all $x$}}_{\forall x^\sigma}\text{ }\underbrace{\text{exists a $y$}}_{\exists y^\tau}\text{ with $R(x,y)$, }\underbrace{\text{then}}_\to\text{ }\underbrace{\text{there is a function $f$}}_{\exists f^{\sigma\to\tau}}\text{ so that }\underbrace{\text{for all $x$}}_{\forall x^\sigma}\text{ holds $R(x,f(x))$".}$$

I skipped over the $\sigma$ and $\tau$ superscripts, as they indicate types and are not of primary importance here.

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