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'pointed set'
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| Title of object: |
pointed set |
| Canonical Name: |
PointedSet |
| Type: |
Definition |
| Created on: |
2006-05-29 20:38:58 |
| Modified on: |
2006-05-30 00:26:06 |
| Classification: |
msc:03E20 |
| Defines: |
basepoint, pointed subset |
Revision comment (for changes between this and next version):
| Changes for correction #7981 ('Second sentence'). |
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Content:
\subsection{Definition}
A \emph{pointed set} is an ordered pair $(A,a)$ such that $A$ is a set and $a\in A$. $a$ is called the \emph{basepoint} of $(A,a)$. At first glance, it seems appropriate enough to call any non-empty set a pointed set. However, the basepoint plays an important role in that if we select a different element $a^{\prime}\in A$, the ordered pair $(A,a^{\prime})$ forms a different pointed set from $(A,a)$. In fact, given any non-empty set $A$ with $n$ elements, $n$ pointed sets can be formed from $A$.
A function $f$ between two pointed sets $(A,a)$ and $(B,b)$ is just a function from $A$ to $B$ such that $f(a)=b$. Whereas there are $|B|^{\mid A\mid}$ functions from $A$ to $B$, only $|B|^{\mid A\mid-1}$ of them are from $(A,a)$ to $(B,b)$.
Pointed sets are mainly used as illustrative examples in the study of universal algebra as algebras with a single constant operator. This operator takes every element in the algebra to a unique constant, which is clearly the basepoint in our definition above. Any \PMlinkname{homomorphism}{HomomorphismBetweenAlgebraicSystems} between two algebras preserves basepoints (taking the basepoint of the domain algebra to the basepoint of the codomain algebra).
\subsection{Creation of Pointed Sets from Existing Ones}
\textbf{Pointed Subsets}. Given a pointed set $(A,a)$, a pointed subset of $(A,a)$ is an ordered pair $(A^{\prime},a)$, where $A^{\prime}$ is a subset of $A$. A pointed subset is clearly a pointed set.
\textbf{Products of Pointed Sets}. Given two pointed sets $(A,a)$ and $(B,b)$, their product is defined to be the ordered pair $(A\times B,(a,b))$. More generally, given a family of pointed sets $(A_i,a_i)$ indexed by $I$, we can form their cartesian product to be the ordered pair $(\prod A_i, (a_i))$. Both the finite and the arbitrary cases produce pointed sets.
\textbf{Quotients}. Given a pointed set $(A,a)$ and an equivalence relation $R$ defined on $A$. For each $x\in A$, define $\overline{x}:=\lbrace y\in A \mid y R x\rbrace$. Then $A/R:=\lbrace \overline{x}\mid x\in A\rbrace$ is a subset of the power set $2^A$ of $A$, called the quotient of $A$ by $R$. Then $(A/R,\overline{a})$ is a pointed set. |
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