10.6: Activities
- Page ID
- 91922
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Suppose \(n\) is a fixed but unknown positive integer, and let \(D: \mathbb{R} \rightarrow \mathbb{R}^n\) represent the function defined by \(D(x) = (x,x, \ldots ,x)\text{.}\)
Write a set definition in Candidate-condition notation for the image set \(D(\mathbb{R})\text{.}\) Then do the same for the graph \(\Delta (D)\text{.}\)
- Devise an example of a function \(\mathbb{N} \to \mathbb{N}\) that is bijective.
- Devise an example of a function \(\mathbb{N} \to \mathbb{N}\) that is injective but not surjective.
- Devise an example of a function \(\mathbb{N} \to \mathbb{N}\) that is surjective but not injective.
Note that when you define a function, you don't necessarily have to give an input-output formula — you can also use a table of input-output values or just a description in words (i.e. an algorithm) of how an output is to be produced from an input.
For each function \(f: A \rightarrow B\) defined below, carry out the following.
- Decide whether the function is injective. Use the Injective Function Test to verify your answer.
- Determine some pattern that all elements of the image \(f(B)\) have in common. That is, if you were handed an arbitrary element of the codomain \(B\text{,}\) describe what property or properties you would use to determine whether it was in the subset \(f(A) \subseteq B\text{.}\)
- Decide whether the function is surjective. Use the Surjective Function Test to verify your answer.
- \(\Sigma = \{0,1\}\text{,}\) \(c: \Sigma^{\ast}\rightarrow \Sigma^{\ast}\) is the bitwise complement function: for input word \(w\text{,}\) the output word \(c(w)\) is the word of the same length as \(w\) but with a \(0\) at every position that \(w\) has a \(1\text{,}\) and a \(1\) at every position that \(w\) has a \(0\text{.}\)
- \(f: \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}\text{,}\) \(f(x) = (x+1,x-1)\text{.}\)
- \(A = \mathscr{P}\mathbb{N} \setminus \{\emptyset\}\text{,}\) \(m: A \rightarrow \mathbb{N} \text{,}\) \(m(X) = \) the smallest number in \(X\text{.}\)
Consider \(\Sigma = \{0,1\}\text{,}\) and recall that for \(n \in \mathbb{N}\text{,}\) \(\Sigma^{\ast}_n\) is the subset of \(\Sigma^{\ast}\) consisting of all words from the alphabet \(\Sigma\) with length \(n\text{.}\) Suppose \(A = \{a_1,a_2,\ldots ,a_n\}\) is a set with \(n\) distinct elements. Construct a bijection \(\mathscr{P}(A) \to \Sigma^{\ast}_n\text{.}\)
When attempting this activity, remember that when you define a function you don't necessarily have to give an input-output formula — you can also use a description in words (i.e. an algorithm) of how an output is to be produced from an input.
Suppose \(A\) is a set that definitely does not contain any cats, and let
\begin{equation*} f: \mathscr{P}(A) \rightarrow \mathscr{P}(A \cup \{\text{Grumpy Cat}\}) \end{equation*}
represent the function defined by
\begin{equation*} f(X) = X \cup \{\text{Grumpy Cat}\} \text{.} \end{equation*}
- Verify that \(f\) is injective.
- Verify that \(f\) is not surjective.
- Describe specifically how to make \(f\) bijective by restricting the codomain.
- As all bijective functions are invertible, the bijective version of \(f\) from Task c has an inverse \(\inv{f}\text{.}\) Describe this inverse by specifying its
- domain,
- codomain, and
- input-output rule.
Let \(\ell: \Sigma^{\ast} \rightarrow \mathbb{N}\) represent the length function, using alphabet is \(\Sigma = \{\alpha,\omega\}\text{.}\)
- Compute \(\ell(\ell ^{-1} (B))\) for \(B = \{1,10,100\}\text{.}\)
- How many elements are there in \(\ell^{-1}(\ell(A))\) for \(A = \{\alpha\alpha, \alpha\omega, \omega\omega\alpha\omega \}\text{?}\)
Suppose \(f: A \rightarrow B\) is a function, and \(B_1,B_2\) are subsets of \(B\text{.}\)
- Draw a Venn diagram illustrating that
\begin{equation*} f^{-1} ( B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1} (B_2) \text{.} \end{equation*}
Include all of the sets
\begin{gather*} A, \;\; B, \;\; B_1, \;\; B_2, \;\; B_1 \cap B_2, \;\; f^{-1}(B_1), \;\; f^{-1} (B_2),\\ f^{-1}(B_1) \cap f^{-1} (B_2), \;\; \text{ and } \;\; f^{-1} ( B_1 \cap B_2) \end{gather*}
in your diagram.
- Formally prove that \(f^{-1} ( B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)\text{,}\) using the Test for Set Equality.
(Note: The parts of this question are independent of one another.)
Suppose \(f: A \rightarrow B\) and \(g: B \rightarrow C\) are functions.
- Argue that if \(f\) and \(g\) are both surjective, then so is \(g \circ f\text{.}\)
- If \(g \circ f\) is surjective, must \(g\) be? Must \(f\) be?
- Argue that if \(f\) and \(g\) are both injective, then so is \(g \circ f\text{.}\)
- If \(g \circ f\) is injective, must \(g\) be? Must \(f\) be?