Limits in $\mathbf {Set}$ I'm a little confused by this example from Leinster's book:

The $x_I$, I suppose, are strictly speaking arrows $x_I: \{\ast\}=1\to D(I)$, and he identifies them with elements of $D(I)$ -- is that right? (If this is so, this part looks okay to me.) But then he says that the set in (5.16) is a limit in Set with projections $p_J(x_I)=x_J$. Aren't $x_J$ themselves projections? The notation $p_J(x_J)$ suggests that $x_J$ are elements of $1=\{\ast\}$ (since $p_J:1=\{\ast\}\to D(J))$ but there is only one element of $1$. 
 A: I think it's easier to see this in a slightly different way: you have a diagram $D:I\to Set$ and want to propose a candidate $(X,\lambda_i)$ for the limit. We must have $Du(\lambda_i x)=\lambda_jx$ for $x\in X$ and $u:i\to j$ and the cone must satisfy the universal property of limit. 
Since for each $i\in I,\ D(i)$ is a set and $\lambda_i$ is an arrow from $X$ to $D(i)$, it makes sense to take $X=\prod_{i\in I}D(i)$ because there are obvious maps from the product to the individual factors: the projections $(x_i)_{i\in I}\mapsto x_i$. These will then be the $(\lambda_i)_{i\in I}$. But there is a condition on the product: it must satisfy $Du(\lambda_i x)=\lambda_jx$. So, we take $X$ to be the set of tuples $(x_i)_{i\in I}$ that satisfy the condition. That is, $X=\{(x_i)_{i\in I}:(\lambda_i x)=\lambda_jx\ \forall\ u:i\to j\}$, and this is Lenister's result. 
Of course, you need to show that if $(Y,\alpha_i)$ is another cone to $D$, then there is a unique $\phi:Y\to X$ such that $\lambda_i\circ \phi=\alpha_i$ for all $i\in I$. But this is easy. Set $\phi(y)=(\alpha_i(y))_{i\in I}$ and check that this is the required arrow. 
A: The first part of your question concerns the isomorphism 
$$ \{ \text{cones on $D$ with vertex $1$} \} \cong \{ (x_I)_{I\in \mathbf I} \mid x_I\in D(I)\, \forall I\in\mathbf I \text{ and } D(u)(x_I) = x_J\, \forall u:I\to J \text { in } \mathbf I\}$$
Indeed, the isomorphism maps a cone $(1,(f_I)_{I\in{\mathbf I}})$, which formally consists of morphisms $f_I : 1 \to D(I)$ for all $I\in\mathbf I$ such that $D(u)\circ f_I = f_J$ for all $u:I\to J$ in $\mathbf I$, to the tuple $(f_I(\ast))_{I\in \mathbf I}$ where $\ast$ denotes the single element of $1$. This tuple is indeed in the right-hand side set because of the conditions imposed on the $f_I$s.
In the second part of your question, you are mistaken in thinking that $p_J: 1 \to D(J$). Let us call $L$ the set on the right-hand side above for simplicity. Leinster is trying to prove that $L$ is the vertex of a limiting cone on $D$, so he needs to provide the "legs" of the cone, for which he proposes $p_J : L \to D(J)$ that maps an element of $L$, necessarily of the form $(x_I)_{I\in\mathbf I}$, to the element $x_J$ in $D(J)$. Notice that he does not claim (and indeed he shouldn't!) that he proved that $L$, together with these $p_J$s, is indeed a limiting cone. He refers to a (future?) exercise for that.
