TITLE: Is there a Yoneda lemma for categories other than Set?
QUESTION [13 upvotes]: The Yoneda lemma says (in my understanding) that instead of studying a category directly, you can study that category's relationships between its relationships into Set. Is the function of Set unique here? Or can other categories like Top do something similar? If so, what are some examples? If not, what is unique about Set? Thank you.

REPLY [7 votes]: There are two ways to interpret your question. Identifying the role of Set in the Yoneda lemma as the category where your categories are enriched in reveals that if you consider $\mathcal V$-enriched categories, than Set there will change to $\mathcal V$ and you get the enriched Yoneda lemma.
Another way of interpreting the question is whether for ordinary categories (or more generally for categories enriched in a fixed given $\mathcal V$, but let's just keep everything simple by considering just ordinary categories) one can replace Set by another category and still make things work nicely. Now, thinking of the Yoneda lemma as turning an abstract category into a category whose objects are sets and whose morphisms are functions, namely as representing an abstract category as a subcategory of something absed on Set (generalising Cayley's representation of an abstract group as a group of permutations) what we are asking is if there is a category $\mathscr C \ne \mathbf {Set}$ such that any abstract (small) category can be represented as a subcategory of something based on $\mathscr C$. (Yes, this is a bit vague.)
The answer is trivially yes, simply by taking any category that contains $\mathbf {Set}$ as a subcategory (so in particular $\mathbf {Top}$ will work). However, there is more to the Yoneda lemma than just the representation. It is really tightly related to representable functors. So, we are asking not just to represent an abstract category, but to represent it in a very particular way. Adding this condition into the question leads to asking: is the codomain of the Yoneda lemma $\mathscr D\to \mathbf{Set}^{\mathscr D^{\mathrm op}}$ unique in some way? The answer is that it is, namely it the cocompletion of $\mathscr D$. So, in this wider context, the base category being $\mathbf {Set}$ is forced.


TITLE: Well-ordered set countability
QUESTION [6 upvotes]: I know there are well-ordered sets that are not countable.
Suppose you are given an uncountable, well-ordered set $S$.
Isn't it possible to provide a bijection $f:\mathbb{N} \rightarrow S$ as following?
$S$ is well-ordered, so it has the smallest element, say $s_1$. $S \setminus$ {$s_1$} is also well-ordered, so there is the next smallest element, $s_2$, Similarly, there is the next smallest element $s_3$, and so on.
Continuing like this, define $f(i) = s_i$.
I know there is something wrong with this, but I cannot really see why...

REPLY [4 votes]: Even if you made a mistake about surjectivity, as pointed out by 5xum and alex, you did a good thing: you proved that
given any two well-ordered sets, one is always the intial segment of the other! (which is, of course, true. The proof needs transfinite induction in general, but works exactly as your attempt of building the bijection)


TITLE: how many ways of distributing marbles
QUESTION [0 upvotes]: Looking through my old statistic books and came upon a very simple question, but I can't just remember the solution.
5 Red Marbles and 5 Blue Marbles. 10 People. How many ways can these marbles be distributed?
the first person has a choice of either red or blue, so it'd be (10C5 or 10C5. The second person either has 5 blue 4 red to choose from or 4 red 5 blue to choose from...and im confused.
How do i solve this? I feel like it's 5! * 5! but not sure..

REPLY [1 votes]: Let the people be indexed as $1,2,\dots,10$. If you would have 10 marbles of 10 different colors, the number of variants would be $10!$ since the first selects from 10, the second from 9, etc.
If you assume that $5!$ of these colors are red and all red marbles are equivalent, you have to divide the number of variants by $5!$ since their order is ignored. Similarly for blue marbles. Thus, the number of possible assignments is
$$
\frac{10!}{5!5!} =252
$$


text:
{\bf Problem.} Let $t(x) = 3x-8$ and $s(t(x)) = x^2 + 3x - 2$.  Find $s(1)$.
{\bf Level.} Level 5
{\bf Type.} Algebra
{\bf Solution.} We don't know $s(x)$, so we don't have an expression we can simply stick $1$ in to get an answer. We do, however, know that $s(t(x)) = x^2 +3x-2$.  So, if we can figure out what to put into $t(x)$ such that $1$ is output, we can use our expression for $s(t(x))$ to find $s(1)$.

If $t(x) = 1$, then $3x-8=1$, which gives $x =3$, so $t(3)=1$.  Therefore, we have $s(t(3)) = s(1)$.  But we also know that $s(t(x)) = x^2 + 3x-2$, so $s(t(3)) = 3^2 +3(3) -2 = \boxed{16}$.

set_name: MATH


TITLE: Proof that ${m+n \choose m} = {m+n \choose n}$
QUESTION [3 upvotes]: This proof follows quite easily using the factorial definition of ${n \choose k}$,
$${m+n \choose m} = \frac{(m+n)!}{m!((m+n)-m)!} = \frac{(m+n)!}{m!n!} = \frac{(m+n)!}{((m+n)-n)!n!} = {m+n \choose n}$$
However I am aware that ${n \choose k}$ is not thought as $\frac{n!}{k!(n-k)!}$ but rather the number of k element subsets of n, and can be proven to equate the the latter formula.
So, my question? What is another proof of this equation using the standard 'thought of' definition of ${n \choose k}$. How can this be proven in terms of subsets? Or simply, is there another way to prove the equation?

REPLY [0 votes]: If you want to form a group of size $m$ out of $m + n$ people, you can either pick the $m$ people in the group or you can pick the $n$ people not in the group, the outcome would be the same.