# Set Exponentiation: Is Y always disjoint from Y^X? [closed]

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?

-

## closed as off topic by Dan Petersen, Kevin Ventullo, quid, Martin Brandenburg, Andrés E. CaicedoNov 17 '12 at 15:54

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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 '12 at 9:09
Yes. Why? Mathematicians overload symbols all the time. – Qiaochu Yuan Nov 17 '12 at 10:08
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 '12 at 11:13
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 '12 at 11:29
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 '12 at 22:14

"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."
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 '12 at 0:06
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 '12 at 0:16
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 '12 at 11:59