|
24 | 24 |
|
25 | 25 | \begin{center} |
26 | 26 | \begin{tikzpicture}[nodes={draw, circle}, -] |
27 | | -\node{r} [grow'=up] |
28 | | - child { node {a} |
29 | | - child { node {c} } |
30 | | - child { node {d} } |
31 | | - child { node {e} } |
| 27 | +\node{$r$} [grow'=up] |
| 28 | + child { node {$a$} |
| 29 | + child { node {$c$} } |
| 30 | + child { node {$d$} } |
| 31 | + child { node {$e$} } |
32 | 32 | } |
33 | | - child { node {b} }; |
| 33 | + child { node {$b$} }; |
34 | 34 | \end{tikzpicture} |
35 | 35 | \end{center} |
36 | 36 |
|
37 | 37 | The lowermost node~$r$ is the root. Every node other than $r$ has |
38 | 38 | exactly one parent node immediately below it. We can think of the relation |
39 | | -a node~$x$ stands in to a node~$y$ if $y$ can be reached from $x$ by |
40 | | -following edges upwards as $x$ being an ancestor of~$y$. |
| 39 | +a node~$x$ stands in to a node~$y$ if $y$ can be reached from~$x$ by |
| 40 | +following edges upwards as $x$ being an \emph{ancestor} of~$y$. |
41 | 41 |
|
42 | 42 | The ancestor relation in a tree is a strict partial order. This |
43 | 43 | motivates the set-theoretic definition. To state it we need two |
|
69 | 69 | \end{prop} |
70 | 70 |
|
71 | 71 | \begin{proof} |
72 | | - Suppose $y < x$ and $y' < x$ and $y \neq y$. Then both $\{y, |
73 | | - y'\} \subseteq \Setabs{z}{z<x}$. Since $\Setabs{z}{z<x}$ is |
74 | | - well-ordered by~$\le$, it has a minimal element, which obviously |
75 | | - must be either $y$ or~$y'$. So either $y \le y'$ or $y' \le y$. We |
76 | | - assumed that $y \neq y'$, so actually either $y < y'$ or $y' < y$. |
77 | | - Since we assumed that $y < x$ and $y' < x$, we furthermore have that |
78 | | - either $y < y' < x$ or $y' < y < x$. So $y$ and $y'$ cannot both be |
| 72 | + Suppose $y_1 < x$ and $y_2 < x$ and $y_1 \neq y_2$. Then $\{y_1, |
| 73 | + y_2\} \subseteq \Setabs{z}{z<x}$. Since $\Setabs{z}{z<x}$ is |
| 74 | + well-ordered by~$\le$, its subset $\{y_1, y_2\}$ has a minimal |
| 75 | + element, which obviously must be either $y_1$ or~$y_2$. So either |
| 76 | + $y_1 \le y_2$ or $y_2 \le y_1$. We assumed that $y_1 \neq y_2$, so |
| 77 | + actually either $y_1 < y_2$ or $y_2 < y_1$. Since we assumed that |
| 78 | + $y_1 < x$ and $y_2 < x$, we furthermore have that either $y_1 < y_2 |
| 79 | + < x$ or $y_2 < y_1 < x$. So $y_1$ and $y_2$ cannot both be |
79 | 80 | predecessors of~$x$. |
80 | 81 | \end{proof} |
81 | 82 |
|
|
110 | 111 | Slightly more generally, the set of finite sequences of natural |
111 | 112 | numbers~$\Nat^*$ with the extension relation~$\sqsubseteq$ is also a |
112 | 113 | tree. It is obviously not finitely branching: every $s \in \Nat^*$ has |
113 | | -infinitely many successurs~$sn$, one for every $n \in \Nat$. Every $A |
| 114 | +infinitely many successors~$sn$, one for every $n \in \Nat$. Every $A |
114 | 115 | \subseteq \Nat^*$ which is closed under~$\sqsubseteq$ is a |
115 | 116 | \emph{subtree} of~$\Nat^*$. (That is, $A$ is such that if $s \in A$ |
116 | 117 | and $s' \sqsubseteq s$, then also $s' \in A$.) All finite trees can be |
|
0 commit comments