# When is the class of functions between sets a set?

I'm reading the paper 'The big fundamental group, big Hawaiian earrings and the big free groups'. The authors state that the class of homotopy equivalences of loops in the space he dubs as the big Hawaiian earrings is not a set. I'm not really certain if what he is saying is true.

I think my question amounts to asking is the class of functions from a set to another set not always a set?

The construction he gives of the big Hawaiian earrings (for ordinal $\omega$) is as the one point compactification of the set of disjoint open intervals with cardinality $\omega$. He then maintains that if $\omega$ has cardinality greater than $\aleph_0$ the resulting class of equivalence classes of homotopic loops is not a group since it's not a set. Is this true? I thought the Hom class of maps between all sets was a set. I'd appreciate any insight, thanks.

• Read the paper more carefully. They are working with "big" loops, which are maps from "big" intervals into the space. Since there is a proper class of "big" intervals there is a proper class of "big" loops (before taking a quotient). – Nik Weaver Apr 23 '13 at 4:58
• @Devin - the axioms of set theory - and you can take this to be ZF(C) or ETCS or SEAR or NBG(C) - imply that the collection of functions between two sets is a set. Only in NF might you find that they form a proper class, but NF (and variants) are not usually taken as foundations for 'ordinary mathematics'. The author of the paper you link considers not just the class of functions from one set to another, but the class of functions $L \to X$ where $L$ is allowed to vary over elements of a proper class, namely the class of 'long lines' of arbitrary cardinality. – David Roberts Apr 23 '13 at 5:04
• However, to quote the paper, "We will be saved from set-theoretic difﬁculties only by the remarkable fact that all maps deﬁned on big intervals into a ﬁxed spaceXfactor through maps deﬁned on relatively small big intervals." This means they only consider in the end a set of maps. And by the way, you can have groups which are proper-class sized. Take, for instance, the additive group underlying the surreal numbers. Or the class of functions $Ord \to\mathbb{Z}$ with pointwise addition. – David Roberts Apr 23 '13 at 5:06
• (The quote only applies to Hausforff spaces, but they are the only ones the authors consider) – David Roberts Apr 23 '13 at 5:09
• Conway uses capital letters. A Group is what would be a group, except the elements constitute a Class but not a set. – Gerald Edgar Apr 23 '13 at 12:52

However, this is not what the authors of the article are considering, rather they are considering (in ordinary classical foundations) a disjoint union $$\coprod_{L\in BigIntervals} Top_\ast(L,X)$$ where each $Top_\ast(L,X)$ is a set. For Hausdorff $X$, the authors state
"We will be saved from set-theoretic difﬁculties only by the remarkable fact that all maps deﬁned on big intervals into a ﬁxed space $X$ factor through maps deﬁned on relatively small big intervals."
Thus they are considering the quotient of the above disjoint union by a relation 'factors through'. That is, for a fixed $X$, there is a set $\{L_\alpha\}_{\alpha\in I}$ of big intervals such that every $f\colon L\to X$ for an arbitrary big interval $L$ factors through some $f_\alpha\colon L_\alpha \to X$, and then $f \sim f_\alpha$. That this isn't a problem is resolved by some variant of Scott's trick, which allows for taking quotients of proper classes, or by arguing directly that the above disjoint union should be replaced by $$\coprod_{L_\alpha\in I} Top_\ast(L_\alpha,X).$$