From c1844fe21f91d32720ab7fb6d9eeab3f2f00ca7b Mon Sep 17 00:00:00 2001 From: Luc Date: Wed, 27 Dec 2023 17:12:52 +0100 Subject: [PATCH] Removed error in formatting proofs. --- docs/en/mathematics/set-theory/maps.md | 20 ++++++++++---------- 1 file changed, 10 insertions(+), 10 deletions(-) diff --git a/docs/en/mathematics/set-theory/maps.md b/docs/en/mathematics/set-theory/maps.md index bd49eca..0f0f5ab 100644 --- a/docs/en/mathematics/set-theory/maps.md +++ b/docs/en/mathematics/set-theory/maps.md @@ -14,10 +14,10 @@ For example, let $f: \mathbb{R} \to \mathbb{R}$ with $f(x) = \sqrt{x}$ for all $
> *Proposition*: let $f: A \to B$ and $g: B \to C$ be maps, then the composition $g$ after $f$: $g \circ f = f;g$ is a map from $A$ to $C$. -> -> ??? note "*Proof*:" -> -> Let $a \in A$ then $g(f(a))$ is an element in $C$ in relation $f;g$ with $a$. If $c \in C$ is an element in $C$ that is in relation $f;g$ with $a$, then there is a $b \in B$ with $afb$ and $bgc$. But then, as $f$ is a map, $b=f(a)$ and as $g$ is a map $c=g(b)$. Hence $c=g(b)=g(f(a))$ is the unique element in $C$ which is in relation $g \circ f$ with $a$. + +??? note "*Proof*:" + + Let $a \in A$ then $g(f(a))$ is an element in $C$ in relation $f;g$ with $a$. If $c \in C$ is an element in $C$ that is in relation $f;g$ with $a$, then there is a $b \in B$ with $afb$ and $bgc$. But then, as $f$ is a map, $b=f(a)$ and as $g$ is a map $c=g(b)$. Hence $c=g(b)=g(f(a))$ is the unique element in $C$ which is in relation $g \circ f$ with $a$.
@@ -40,9 +40,9 @@ If $B'$ is a subet of $B$ then the pre-image of $B'$, denoted by $f^{-1}(B') is > > * If $A' \subseteq A$, then $f^{-1}(f(A')) \supseteq A'$. > * If $B' \subseteq B$, then $f(f^{-1}(B')) \subseteq B'$. -> -> ??? note "*Proof*:" -> -> Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$. -> -> Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$. \ No newline at end of file + +??? note "*Proof*:" + + Let $a' \in A'$, then $f(a') \in f(A')$ and hence $a' \in f^{-1}(f(A'))$. Thus $A' \subseteq f^{-1}(f(A'))$. + + Let $a \in f^{-1}(B')$, then $f(a) \in B'$. Thus $f(f^{-1}(B')) \subseteq B'$. \ No newline at end of file