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If $y \in Y$ and $g \in Y^X$, we often write $y+g$ as shorthand for the map $x \mapsto y+ g(x)$. Similarly if $f \in Y^X$ then $f+g = x \mapsto f(x)+g(x)$. However this presupposes that we can distinguish between an element of $Y$ and an element of $Y^X$. That is, we require these sets be disjoint. Are they?

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 Why do you need to presuppose such a thing? Just declare that $y$ is shorthand for the constant map $X \to Y$ with value $y$, then note that if $Y$ is a monoid then $Y^X$ canonically inherits a monoid structure. None of what I've said depends on questions like whether $Y$ and $Y^X$ are disjoint (and such questions are not even meaningful in the version of set theory in my head). – Qiaochu Yuan Nov 17 at 9:09
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 Yes. Why? Mathematicians overload symbols all the time. – Qiaochu Yuan Nov 17 at 10:08
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 Of course, in general the sets may not be disjoint, so the problem you identify actually occurs. The fact is that $Y$ may have a function from $X$ to (some other part of) $Y$ as an element. For example, consider $Y=\text{HC}$, the set of all hereditary countable sets, and let $X=\omega$; observe in this case that $Y^X\subset Y$, since any function from $\omega\to\text{HC}$ is itself hereditarily countable. Similar examples abound. But meanwhile, this is rarely a problem for mathematical communication, since one can resolve ambiguities in notation by explaining what is meant. – Joel David Hamkins Nov 17 at 11:13
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 Qiaochu, your proposed solution about reconsidering every element of $Y$ to be a constant map from $X$ to $Y$ doesn't actually resolve the ambiguity, in the case that $Y$ itself has that map as a point. In other words, it could be that your new version of $3+g$ is still ambiguous, if the constant $3$ map is also in $Y$ (as it is in my example with HC). That is, are you adding $3$ to each point $g(x)$, or are you adding the constant map $3$ to each point $g(x)$? (Absurd, I know...) – Joel David Hamkins Nov 17 at 11:29
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 I should have written "hereditarily countable" rather than "hereditary countable". And it is fine, Yianni, that you posted my comment as an answer. – Joel David Hamkins Nov 17 at 22:14
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closed as off topic by Dan Petersen, Kevin Ventullo, quid, Martin Brandenburg, Andres Caicedo Nov 17 at 15:54

1 Answer

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As Joel David Hamkins points out, the assertion is false.

"The fact is that $Y$ may have a function from $X$ to (some other part of) $Y$ as an element. For example, consider $Y=\mathrm{HC}$, the set of all hereditarily countable sets, and let $X=\omega$; observe in this case that $Y^X \subset Y$, since any function from $ω \rightarrow HC$ is itself hereditarily countable. Similar examples abound."

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Joel: If $Y^X=Y$, $x \in X$ and $f \in Y$ wouldn't the sequence $f, f(x), f(x)(x), \dots$ contradict foundation? – Ramiro de la Vega Nov 18 at 0:06
Ramiro, yes, I had come to the same conclusion myself, and had just previously deleted my comment. – Joel David Hamkins Nov 18 at 0:13
But meanwhile, what my argument shows is that one can arrange $Y^X=Y\cup Y_0$ for any given nonempty $Y_0$, by starting with $Y_0$, and then adding all functions from $X$ to what you have so far. After $|X|^+$ iterations, you get $Y$ with $Y^X=Y\cup Y_0$. – Joel David Hamkins Nov 18 at 0:16
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 Joel, in your last comment, the equation should be $Y=Y_0\cup Y^X$. That is, $Y_0$ is part of $Y$, not of $Y^X$. – Andreas Blass Nov 19 at 11:59
Andreas, yes, that's right. – Joel David Hamkins Dec 4 at 2:29

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