# Is every set a pointed set?

My question is quite simple, It seems every non-empty set is a pointed set, only we have to do is choice some element to be the distinguished element, am I right? I'm looking for non-empty sets which aren't pointed sets.

EDIT

I think I didn't express myself clear, my question is can every set be transformed into a pointed set, just picking up a point to be the distinguished point?

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If $A$ is a (nonempty) set, then $A$ is not a pointed set. However, if $a\in A$, then $\langle A,a\rangle$ is a pointed set. So there is little sense (not to say: no point) in talking about a set being pointed or not before you pick a point. – Hagen von Eitzen Apr 6 '14 at 12:23
Can you define pointed set? – Felipe Apr 6 '14 at 12:23
– user42912 Apr 6 '14 at 12:42

A non-empty set is not a pointed set until you actually pick a point in it. The pointed set is the combination of a set and a particular element in it.

So $\{1,2\}$ is not a pointed set, but $\langle \{1,2\},1\rangle$ and $\langle\{1,2\},2\rangle$ are two different pointed sets that can be made from it.

And if somebody talks about "pointed sets" rather than just "nonempty sets", it must be because exactly that difference is important to them in the context!

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I think I didn't express myself clear, my question is can every set be transform into a pointed set, just picking up a point to be the distinguished point? – user42912 Apr 6 '14 at 12:45
@user42912: Yes. – Henning Makholm Apr 6 '14 at 12:47

There is a functor from the category of pointed sets to the category of sets, that is, forgetting the chosen point. However there's no “canonical” functor backwards unless you define it by choosing anything which is not a set and add it.

Actually the category of pointed sets is equivalent to the category of sets with partial maps, which is not equivalent to the category of sets with (everywhere defined) maps.

What's a partial map $f$ from $A$ to $B$? It's a quadruple $(A,A',f,B)$ where $A'\subseteq A$ and $f$ is a (usual) map $f\colon A'\to B$.

The equivalence is readily established: map every pointed set $(A,a)$ to $F(A)=A\setminus\{a\}$ and a map of pointed set $f\colon (A,a)\to (B,b)$ to the partial map $(F(A),A'=A\setminus f^{-1}(\{b\}),f|_{A'},F(B))$.

This is readily seen to be a functor from $\mathbf{PSet}\to\mathbf{Set}$. The “inverse” functor is obtained by defining $G(A)=A\cup\{*\}$, where $*$ is a formal symbol equal only to itself and not belonging to any set; if $p=(A,A',f,B)$ is a partial map from $A$ to $B$, define $G(p)$ to be the map $G(p)\colon G(A)\to G(B)$ such that $$G(p)\colon x\mapsto \begin{cases} f(x)&\text{if x\in A'},\\ * & \text{if x\in A\setminus A'},\\ * & \text{if x={*}}. \end{cases}$$

Verifying that $G$ is a functor and that $(F,G)$ is an equivalence of categories is easy.

So, no, a non empty set is not a pointed set. You can choose a point so that it becomes a set, but nothing guarantees that when you have a map $f\colon A\to B$ of non empty sets, the choices you do in $A$ and $B$ makes this map into a map of pointed sets.

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user 42912: this example is in Awodey too. You will find it soon. – magma Apr 17 '14 at 9:47

Every single nonempty set can be made into a pointed set by choosing out an arbitrary element. A collection of infinite number of nonempty set, however, need not be capable of all being made into pointed set, if Axiom of Choice is not accepted.

Consider an index set $I$ that is infinite. Let $A_{i}$ be a collection of nonempty set, and let $B_{i}$ be a collection of pointed set. Assume an universe where Axiom of Choice is not true. Then:

$\prod_{i\in I}A_{i}$ can be an empty set. But $\prod_{i\in I}B_{i}$, as a set, is nonempty; in fact it contains at least one point, with each coordinate being the distinguished point in each point set.

So there are huge differences. Choosing can be very hard indeed.

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Interesting example, thank you very much – user42912 Apr 10 '14 at 19:36