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packages/macros.sty : Added suchthat operator && typos fixes

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saundersp
2025-03-11 14:43:31 +01:00
parent b23bc3bd4d
commit 52fd077cb5
7 changed files with 31 additions and 30 deletions
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14
contents/set_theory.tex
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@ -46,7 +46,7 @@ $(a, b)_K := \{\{a\}, \{a, b\}\}$
Unite all elements of two given sets into one.
\begin{definition_sq} \label{definition:set_union}
$A \union B := \{x \mid (x \in A \lor x \in B)\}$
$A \union B := \{x \suchthat (x \in A \lor x \in B)\}$
\end{definition_sq}
Pour des ensembles finis : $\forall (E, F) \in \Cat(\Set)^2, \card{E \union F} = \card{E} + \card{F} - \card{E \intersection F}$
@ -89,9 +89,9 @@ The axiom of choice implies the law of excluding middle.
Assume that $0 \ne 1$ (or any two elements that are not equal), Let $\Omega := \{0, 1\}$, $p \in \mathbf{Prop}$
$A := \{ x \in \Omega \mid x = 0 \lor p \}$
$A := \{ x \in \Omega \suchthat x = 0 \lor p \}$
$B := \{ y \in \Omega \mid y = 1 \lor p \}$
$B := \{ y \in \Omega \suchthat y = 1 \lor p \}$
$\implies 0 \in A \land 1 \in B$
@ -115,10 +115,10 @@ So by proof by cases $(p \lor \lnot p)$ which is the law of excluded middle \ref
Unite all common elements of two given sets into one.
\begin{definition_sq} \label{definition:set_intersection}
$A \intersection B := \{x \mid (x \in A \land x \in B)\}$
$A \intersection B := \{x \suchthat (x \in A \land x \in B)\}$
\end{definition_sq}
Pour des ensembles finis : $\forall E,F \in \Cat(\Set), \card{E \intersection F} = \card{E} - \card{F} + \card{E \union F}$
Pour des ensembles finis : $\forall (E, F) \in \Set^2, \card{E \intersection F} = \card{E} - \card{F} + \card{E \union F}$
Example :
@ -136,10 +136,10 @@ $A \intersection B = \{c_0, \cdots, c_n\}$
Exclude elements of a set from a set
\begin{definition_sq} \label{definition:set_difference}
$A \setminus B := \{x \mid (x \in A \land x \notin B)\}$
$A \setminus B := \{x \suchthat (x \in A \land x \notin B)\}$
\end{definition_sq}
Pour des ensembles finis : $\forall E,F \in \Cat(\Set), \card{E \setminus F} = \card{E} - \card{E \intersection F}$
Pour des ensembles finis : $\forall (E, F) \in \Set^2, \card{E \setminus F} = \card{E} - \card{E \intersection F}$
\langsection{Fonction}{Function}