Here, we show demonstrate the general format of a theorem
Figure is next
Figure3.2.1Venn Diagram
A theorem is contained within a \(\texttt{theorem}\) tag, normally with two inner tags: \(\texttt{statement}\) and \(\texttt{proof}\). By default, the proof will be a knowl.
Open question: I've used the LaTeX \blacksquare at the end of the proof. Is there a way to automatically have an "end of proof" symbol appear at the end of proofs? In a knowl it might not be needed?
Theorem3.2.2Charactorization of bijective functions
Let \(f: A \rightarrow A\). There exists an inverse of \(f\) if and only if \(f\) is a bijection; i. e. \(f\) is one-to-one and onto.
(\(\Rightarrow\)) In this half of the proof, assume that \(f^{-1}\) exists and we must prove that f
is one-to-one and onto. To do so, it is convenient for us to use the relation notation, where \(f(s)=t\)
is equivalent to \((s,t)\in f\).
To prove that \(f\) is one-to-one, assume that \(f(a)=f(b) = c\). Alternatively, that means \((a, c)\)
and \((b, c)\) are elements of \(f\). We must show that \(a=b\).
\begin{equation*}(a, b), (c, b) \in f \Rightarrow (c, a), (c,b) \in f^{-1}\end{equation*}
By the fact that \(f^{-1}\) is a function and \(c\) cannot have two images, \(a\) and \(b\) must be equal, so \(f\) is one-to-one.
Next, to prove that \(f\) is onto, observe that for \(f^{-1}\) to be a function, it must use all of its domain, namely \(A\). Let \(b\) be any element of \(A\). Then \(b\) has an image under \(f^{-1}\), \(f^{-1}(b)\). Another way of writing this is \((b,f^{-1}(b)) \in f^{-1}\), By the definition of the inverse, this is equivalent to \((f^{-1}(b),b) \in f\). Hence, \(b\) is in the range of \(f\). Since \(b\) was chosen arbitrarily, this shows that the range of \(f\) must be all of \(A\) and so \(f\) is onto.
\(\Rightarrow\) Assume \(f\) is one-to-one and onto and we are to prove \(f^{-1}\) exists. We leave this half of the proof to the reader. \(\blacksquare\)
