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I was curious whether there is a standard reference for the correct English pronunciation of general mathematical symbols. In particular, I am wondering how to pronounce the symbol "$\in$" correctly, or at least what is most common.

In mathematical texts, I have seen phrases like "Let $x \in X$ such that" and "Let $x \in X$ be such that". In the first phrase $\in$ seems to be pronounced as "be an element of", whereas the correct pronunciation in the second phrase is just "in".

In the excellent survey How to write mathematics?, Paul Halmos writes that he prefers to read "$\in"$ as "is in" rather than the preposition "in". This still seems to be very personal, so I am curious whether there is a standard nowadays.

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    $\begingroup$ The $\LaTeX$ code for $\in$ is \in. This gives a clue. $\endgroup$ – Clement C. Dec 29 '16 at 17:14
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    $\begingroup$ Whatever makes sense linguistically, tbh. If we write "Let $x \in X$", we'd read "Let $x$ be an element of $X$", whereas if we write "Suppose $x \in X$", we'd read "Suppose $x$ is an element of $X$" or even "Suppose $x$ is in $X$". The symbol "$\in$" is a placeholder for an idea; it doesn't particularly matter how the idea is expressed as long as it is. $\endgroup$ – Kaj Hansen Dec 29 '16 at 17:30
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    $\begingroup$ @ClementC. Then should $\oint$ be pronounced as "oint"? $\endgroup$ – Henricus V. Dec 29 '16 at 17:42
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    $\begingroup$ @HenryW. Devil's advocate? When the grammar is clearly linked to the semantics (\in, \to, \subset, \notin, \mapsto), taking it as a clue for the pronunciation makes sense. (That's also why using \to when writing $f\colon X\to Y$, instead of the identically-looking \rightarrow, is a good idea.) $\endgroup$ – Clement C. Dec 29 '16 at 17:44
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There are a few idiomatic usages. I'll copy from a randomly-selected textbook on my shelf. This is from a proof of Cantor's Theorem: For every set $A$, $A$ and its power set $\mathscr{P}(A)$ do not have the same size:

Let $f \colon A \to \mathscr{P}(A)$ be given. We will show that $f$ is not onto. Define $$ B = \left\{ x\in A \mid x \notin f(x) \right\}$$ Since $B \subseteq A$, this means that $B \in \mathscr{P}(A)$ [1], so $B$ is a member of the codomain of $f$. To see that $f(x) \neq B$ for all $x \in A$ [2], we imagine what would happen if there were a $b\in A$ with $f(b) = B$. A contradiction then arises when considering the question “Is $b\in B$? [3]” ...

From Discrete Mathematics: Mathematical Reasoning and Proof with Puzzles, Patterns, and Games, by Douglas E. Ensley and J. Winston Cawley.

In the first note “this means that $B\in\mathscr{P}(A)$,” we can read $\in$ as “is in.” It would be equivalent to say “this means that $B$ is in $\mathscr{P}(A)$.” The main idea of the sentence is that the relation holds.

In the second, “$f(x) \neq B$ for all $x \in A$”, the $\in$ can be read as just “in.” As in, “$f(x) \neq B$ for all $x$ in $A$.” To put “is” in there would insert an extra verb. The relation $x\in A$ is not the main idea here; it's parenthetical to $f(x) \neq B$.

In the third, the relation is still the main idea of the clause, but to make the question start with a verb (standard written English), we write is and read “in” for $\in$.

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