\magnification 1200 \nopagenumbers \parindent 0pt \parskip=10pt \voffset 20pt Math 2418 Linear Algebra Exam \#3~~~ Part A \vskip -10pt October 20, 2002 \hskip 2in SHOW ALL WORK [7]~~ 1.) Show that the inner product $<(u_1, u_2), (v_1, v_2)> ~~=~ 3u_1v_1 + 4u_2v_2$ satisfies the additivity axiom of an inner product: $<{\bf u} + {\bf v}, {\bf w}> ~~=~ < {\bf u}, {\bf w}> + <{\bf v}, {\bf w}>$ $<{\bf u} + {\bf v}, {\bf w}> ~~=~<(u_1, u_2) + (v_1, v_2), (w_1, w_2)> ~=~<(u_1 + v_1, u_2+v_2), (w_1, w_2)> ~=~ 3(u_1 + v_1)w_1 + 4(u_2 + v_2)w_2 ~=~ 3u_1w_1 + 3v_1w_1 + 4u_2w_2 + 4v_2w_2$ $< {\bf u}, {\bf w}> + <{\bf v}, {\bf w}> ~=~ 3u_1w_1 + 4u_2w_2 + 3v_1w_1 + 4v_2w_2~=~ 3u_1w_1 + 3v_1w_1 + 4u_2w_2 + 4v_2w_2$ Hence, $<{\bf u} + {\bf v}, {\bf w}> ~~=~ < {\bf u}, {\bf w}> + <{\bf v}, {\bf w}>$ [7]~~ 2.) Show that $<(u_1, u_2), (v_1, v_2)> ~~=~ 3u_2v_1 + 4u_1v_2$ is not an inner product by giving a specific example (using only real numbers) and showing that the inner product definition or a consequence of it does not hold. $<(1, 0), (0, 1)> = 3(0)(0) + 4(1)(1) = 4$ $<(0, 1), (1, 0)> = 3(1)(1) + 4(0)(0) = 4$ Thus, $<(1, 0), (0, 1)> \not= <(0, 1), (1, 0)>$ Alternate answer: $<(1, 0), (1, 0)> = 3(0)(1) + 4(0)(1) = 0$ but $(1, 0) \not= (0, 0)$. Note there are many alternate answers. \eject Math 2418 Linear Algebra Exam \#3~~~ Part B \vskip -10pt October 20, 2002 \hskip 2in SHOW ALL WORK [19]~ 3.) The matrix $A = \left[\matrix{ 0 & 0 & 0 & 0 \cr 0 & 0 & 0 & 0 \cr 0 & 0 & 1 & 4 \cr 0 & 0 & 2 & 8 \cr }\right]$ has two eigenvalues. Find these eigenvalues and bases for the corresponding eigenspaces (Note: this problem continues on the next page). $\lambda I - A = \left|\matrix{ \lambda & 0 & 0 & 0 \cr 0 & \lambda & 0 & 0 \cr 0 & 0 & \lambda -1 & -4 \cr 0 & 0 & -2 & \lambda -8 \cr }\right| = \lambda^2[(\lambda-1)(\lambda - 8) - 8] = \lambda^2(\lambda^2 - 9\lambda) = \lambda^3(\lambda - 9)$ Thus $\lambda = 0$ is an eigenvalue of $A$ with algebraic multiplicity 3 and $\lambda = 9$ is an eigenvalue of $A$ with algebraic multiplicity 1. $\lambda = 0$: Solve $(\lambda I - A){\bf x} = {\bf 0}$ $\lambda I - A= 0(I) - A = A $ $\left[\matrix{ 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0\cr 0 & 0 & -1 & -4 & 0\cr 0 & 0 & -2 & -8 & 0\cr }\right] \rightarrow \left[\matrix{ 0 & 0 & 1& 4 & 0\cr 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0\cr }\right]$ Thus, $\left[\matrix{ x_1\cr x_2\cr x_3\cr x_4\cr }\right] = \left[\matrix{ r\cr s\cr -4t\cr t\cr }\right] = r\left[\matrix{ 1\cr 0\cr 0\cr 0\cr }\right] + s\left[\matrix{ 0\cr 1\cr 0\cr 0\cr }\right] + t\left[\matrix{ 0\cr 0\cr -4\cr 1\cr }\right] $ An eigenvalue of $A$ is $\underline{~0~}$ A basis corresponding to this eigenvalue is $\left[\matrix{ 1\cr 0\cr 0\cr 0\cr }\right], \left[\matrix{ 0\cr 1\cr 0\cr 0\cr }\right], \left[\matrix{ 0\cr 0\cr -4\cr 1\cr }\right] $ List 5 eigenvectors corresponding to this eigenvalue. $\left[\matrix{ 1\cr 0\cr 0\cr 0\cr }\right], \left[\matrix{ 0\cr 1\cr 0\cr 0\cr }\right], \left[\matrix{ 0\cr 0\cr -4\cr 1\cr }\right] $, $\left[\matrix{ 2\cr 0\cr 0\cr 0\cr }\right], \left[\matrix{ 1\cr 1\cr -4\cr 1\cr }\right] $ or any linear combination of the basis vectors except the zero vector (recall that the zero vector is by definition never an eignenvector even though it is always in the eigenspace - since the zero vector is in every vector space). \eject 3 cont.) $\lambda = 0$: Solve $(\lambda I- A){\bf x} = {\bf 0}$ $\lambda I - A = 9(I) - A$ $\left[\matrix{ 9 & 0 & 0 & 0 & 0\cr 0 & 9 & 0 & 0 & 0\cr 0 & 0 & 8 & -4 & 0\cr 0 & 0 & -2 & 1 & 0\cr }\right] \rightarrow \left[\matrix{ 1 & 0 & 0 & 0 & 0\cr 0 & 1 & 0 & 0 & 0\cr 0 & 0 & 1 & -{1 \over 2} & 0\cr 0 & 0 & 0 & 0 & 0\cr }\right] \rightarrow Thus, $\left[\matrix{ x_1\cr x_2\cr x_3\cr x_4\cr }\right] = \left[\matrix{ 0\cr 0\cr {1 \over 2}t\cr t\cr }\right] = t\left[\matrix{ 0\cr 0\cr {1 \over 2}\cr 1\cr }\right]$ A second eigenvalue of $A$ is $\underline{~9~}$ A basis corresponding to this second eigenvalue is (circle your answer) $\left[\matrix{ 0\cr 0\cr {1 \over 2}\cr 1\cr }\right]$ or any multiple of this vector such as $\left[\matrix{ 0\cr 0\cr 1\cr 2\cr }\right]$ \vskip 2.9in \hrule \vskip 10pt 4.) If $C \left[\matrix{ 1 \cr 2 \cr }\right] = \left[\matrix{ 0 \cr 0 \cr }\right]$, $C\left[\matrix{ 0 \cr 0 \cr }\right] = \left[\matrix{ 0 \cr 0 \cr }\right]$, and $C\left[\matrix{ 5 \cr 3 \cr }\right] = \left[\matrix{ -10 \cr ~-6 \cr }\right]$, \vskip 10pt [3]~~ 4a.) The eigenvalue(s) of $C$ is/are $\underline{ ~0, ~-2~ }$ \vskip 30pt [3]~~ 4b.) Three eigenvectors of $C$ are $\underline{\left[\matrix{ 1 \cr 2 \cr }\right], \left[\matrix{ 2 \cr 4 \cr }\right], \left[\matrix{ 3 \cr 6 \cr }\right]}$ Note any multiple of $\left[\matrix{ 1 \cr 2 \cr }\right]$ is an eigenvector of $C$ (corresponding to eigenvalue 0) and any multiple of $\left[\matrix{ 5 \cr 3 \cr }\right]$ is an eigenvector of $C$ (corresponding to eigenvalue -2) \vskip 30pt [2]~~ 4c.) A vector which is not an eigenvector of $C$ is $\underline{\left[\matrix{ 1 \cr 3 \cr }\right]}$ Any vector which is not a multiple of $\left[\matrix{ 1 \cr 2 \cr }\right]$ nor a multiple of $\left[\matrix{ 5 \cr 3 \cr }\right]$ is not an eigenvector of $C$ \vskip 30pt 5.) Circle T for true and F for false. [4]~~ 5a.) Matlab always finds an exactly correct answer when solving $A{\bf x} = {\bf b}$ when the columns of $A$ are linearly independent. \hfill ~~~~~~~~~~F [4]~~ 5b.) If $\lambda = 0$ is an eigenvalue of $A$, then $A$ is NOT invertible. \hfill T ~~~~~~~~~~ \eject [15]~~ 6a.) Find a QR factorization of $A = \left[\matrix{ 2 & 3 \cr 2 & 1 \cr 1 & 4 \cr }\right]$. Gram-Schmidt: $(2, 2, 1) \cdot (3, 1, 4) = 6 + 2 + 4 = 12$ $(2, 2, 1) \cdot (2, 2, 1) = 4 + 4 + 1 = 9$ }$proj_{\left[\matrix{ 2 \cr 2 \cr 1 \cr }\right]} \left[\matrix{ 3 \cr 1 \cr 4 \cr }\right] = {12 \over 9}\left[\matrix{ 2 \cr 2 \cr 1 \cr }\right] = {4 \over 3}\left[\matrix{ 2 \cr 2 \cr 1 \cr }\right] = \left[\matrix{ {8 \over 3} \cr {8 \over 3} \cr {4 \over 3} \cr }\right]$ $\left[\matrix{ 3 \cr 1 \cr 4 \cr }\right] - \left[\matrix{ {8 \over 3} \cr {8 \over 3} \cr {4 \over 3} \cr }\right] = \left[\matrix{ {1 \over 3} \cr -{5 \over 3} \cr {8 \over 3} \cr }\right]$ Check: \left[\matrix{ 2 \cr 2 \cr 1 \cr }\right] \cdot \left[\matrix{ {1 \over 3} \cr -{5 \over 3} \cr {8 \over 3} \cr }\right] = {2\over 3} -{10 \over 3} + {8 \over 3} = 0$ Normalize: {1 \over \sqrt 9}\left[\matrix{ 2 \cr 2 \cr 1 \cr }\right] =\left[\matrix{ {2 \over 3} \cr {2 \over 3} \cr {1 \over 3} \cr }\right] \left[\matrix{ {1 \over 3} \cr -{5 \over 3} \cr {8 \over 3} \cr }\right] \cdot \left[\matrix{ {1 \over 3} \cr -{5 \over 3} \cr {8 \over 3} \cr }\right] = {1 \over 9} + {25 \over 9} + {64 \over 9} = {90 \over 9} = 10 {1 \over \sqrt 10}\left[\matrix{ {1 \over 3} \cr -{5 \over 3} \cr {8 \over 3} \cr }\right] = \left[\matrix{ {1 \over 3\sqrt 10} \cr -{5 \over 3\sqrt 10} \cr {8 \over 3\sqrt 10} \cr }\right] $QR = A$. Hence $R = Q^TQR = Q^TA = \left[\matrix{ {2 \over 3} & {2 \over 3} &{1 \over 3} \cr {1 \over 3\sqrt 10} & -{5 \over 3\sqrt 10} & {8 \over 3\sqrt 10} }\right] \left[\matrix{ 2 & 3 \cr 2 & 1 \cr 1 & 4 \cr }\right] = \left[\matrix{ 3 & 4 \cr 0 & 3 \sqrt 10 \cr }\right]$ Answer 6a.) $\underline{Q = \left[\matrix{ {2 \over 3} & {1 \over 3\sqrt 10} \cr {2 \over 3} & -{5 \over 3\sqrt 10} \cr {1 \over 3} & {8 \over 3\sqrt 10} \cr }\right] } \hskip 0.8in \underline{R = \left[\matrix{ 3 & 4 \cr 0 & 3 \sqrt 10 \cr }\right]}$ [4]~~ 6b.) An orthonormal basis (according to the usual dot product) for the column space of $A$ is \left[\matrix{ {2 \over 3} \cr {2 \over 3} \cr {1 \over 3} \cr }\right] \left[\matrix{ {1 \over 3\sqrt 10} \cr -{5 \over 3\sqrt 10} \cr {8 \over 3\sqrt 10} \cr }\right] [2]~~ 6c.) A basis for the nullspace of $A^T$ is Solve augmented matrix $\left[\matrix{ 2 & 2 & 1 & 0 \cr 3 & 1 & 4 & 0 \cr }\right] \rightarrow \left[\matrix{ 1 & 1 & {1 \over 2} & 0 \cr 3 & 1 & 4 & 0 \cr }\right] \rightarrow \left[\matrix{ 1 & 1 & {1 \over 2} & 0 \cr 0 & -2 & {5 \over 2 & 0 \cr }\right] \rightarrow \left[\matrix{ 1 & 1 & {1 \over 2} & 0 \cr 0 & 1 & -{5 \over 4 & 0 \cr }\right] \rightarrow \left[\matrix{ 1 & 0 & {7 \over 4} & 0 \cr 0 & 1 & -{5 \over 4 & 0 \cr }\right]$ or note nullspace of $A^T$ is orthogonal to the column space of $A$. Since these spaces live in $R^3$ and the column space of $A$ is a 2-dimensional plane, the nullspace of $A^T$ = orthogonal complement of column space of $A$ is a line. A vector which is perpendicular to the column space is $\left|\matrix{ i & j & k \cr 2 & 2 & 1 \cr 3 & 1 & 4 \cr }\right| = (8 - 1, -(8 - 3), 2 - 6) = (7, -5, -4)$ Thus a basis for the nullspace of $A^T $ is (7, -5, -4). \eject [25]~~ 7.) $\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \}$ is an orthogonal set (but not an orthonormal set) given the inner product $ = \int_{-1}^1 fg dx$. Use this inner product to solve the following problems. 7a.) The projection of $3x$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = \underline{{4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2)}$ $\int_{-1}^1(3x)( 1 + 2x + x^2)dx = \int_{-1}^1 (3x + 6x^2 + 3x^3)dx = ({3 \over 2}x^2 + 2x^3 + {3 \over 4}x^4)|_{-1}^1 = 2 - (-2) = 4$. $\int_{-1}^1(3x)( 1 + 2x - 5x^2) = \int_{-1}^1 (3x + 6x^2 - 15x^3)dx = ({3 \over 2}x^2 + 2x^3 - {15 \over 4}x^4)|_{-1}^1 = 2 - (-2) = 4$. Thus, if $||1 + 2x + x^2|| = A$ and $||1 + 2x - 5x^2|| = B$, then the projection of $3x$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = {4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2)$ 7b.) The projection of $15x^2$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = \underline{\hskip 1in}$ $\int_{-1}^1(15x^2)( 1 + 2x + x^2)dx = \int_{-1}^1 (15x^2 + 30x^3 + 15x^4)dx = ({5}x^3 + {30 \over 4}x^4 + {3}x^5)|_{-1}^1 = 5 - (-5) + 3 - (-3) = 16$. $\int_{-1}^1(15x^2)( 1 + 2x - 5x^2)dx = \int_{-1}^1 (15x^2 + 30x^3 - 75x^4)dx = ({5}x^3 + {30 \over 4}x^4 + {15}x^5)|_{-1}^1 = 5 - (-5) + 15 - (-15) = 40$. Thus, if $||1 + 2x + x^2|| = A$ and $||1 + 2x - 5x^2|| = B$, then the projection of $3x$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = {16 \over A^2}(1 + 2x + x^2) + {40 \over B^2}(1 + 2x - 5x^2)$ 7c.) If possible, write the $3x$ as a linear combination of $\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \}$. If not possible, state not possible. Not possible 7d.) If possible, write the $15x^2$ as a linear combination of $\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \}$. If not possible, state not possible. {16 \over A^2}(1 + 2x + x^2) + {40 \over B^2}(1 + 2x - 5x^2)$ 7e.) If $W = span\{1 + 2x + x^2, ~~1 + 2x - 5x^2\}$, find a basis for $W^\perp$. $3x - ({4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2))$ 7f.) Find an orthogonal basis (according to the given inner product) for $P_2$ which contains the polynomials $1 + 2x + x^2$ and $1 + 2x - 5x^2$ $1 + 2x + x^2$, $1 + 2x - 5x^2$, $3x - ({4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2))$ 7g.) Is your answer to 5f an orthogonal basis for $P_2$ according to the inner product $ ~~=~ a_0b_0 + a_1b_1 + a_2b_2$. \end $\int_{-1}^1(1 + 2x + x^2)( 1 + 2x + x^2)dx = \int_{-1}^1 (1 + 4x + 6x^2 + 4x^3 +x^4)dx = x + 2x^2 + 2x^3 + x^4 + {1 \over 5}x^5 |_{-1}^1 = 1 - (-1) + 2 - (-2) + {1 \over 5} - (-{1 \over 5}) = {32 \over 5}$ Thus, $A^2 = {32 \over 5}$ $\int_{-1}^1(1 + 2x - 5x^2)( 1 + 2x - 5x^2)dx = \int_{-1}^1 (1 + 4x - 6x^2 - 20x^3 + 25x^4)dx = x + 2x^2 - 2x^3 - 5x^4 + {5}x^5 |_{-1}^1 = 1 - (-1) - [2 - (-2)] + {5} - (-{5}) = 2 - 4 + 10 = 8$ Thus, $B^2 = {8}$ 7a.) The projection of $3x$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = \underline{{4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2)}$ the projection of $3x$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = {5 \over 8}(1 + 2x + x^2) + {4 \over 8}(1 + 2x - 5x^2) = {9 \over 8} + 18 \over 8 x - {15 \over 8}x^2$ 7b.) The projection of $15x^2$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = \underline{\hskip 1in}$ the projection of $3x$ onto span$\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \} = {16 \over A^2}(1 + x^2) + {40 \over B^2}(1 + 2x - 5x^2) = {5 \over 2}(1 + 2x + x^2) + {5}(1 + 2x - 5x^2) = {15 \over 2} $ 7c.) If possible, write the $3x$ as a linear combination of $\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \}$. If not possible, state not possible. Not possible 7d.) If possible, write the $15x^2$ as a linear combination of $\{ 1 + 2x + x^2, ~~1 + 2x - 5x^2 \}$. If not possible, state not possible. {16 \over A^2}(1 + 2x + x^2) + {40 \over B^2}(1 + 2x - 5x^2)$ 7e.) If $W = span\{1 + 2x + x^2, ~~1 + 2x - 5x^2\}$, find a basis for $W^\perp$. $3x - ({4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2))$ 7f.) Find an orthogonal basis (according to the given inner product) for $P_2$ which contains the polynomials $1 + 2x + x^2$ and $1 + 2x - 5x^2$ $1 + 2x + x^2$, $1 + 2x - 5x^2$, $3x - ({4 \over A^2}(1 + 2x + x^2) + {4 \over B^2}(1 + 2x - 5x^2))$ 7g.) Is your answer to 5f an orthogonal basis for $P_2$ according to the inner product $ ~~=~ a_0b_0 + a_1b_1 + a_2b_2$. Topology (many answers modified from latexed HW of Colin B. McKinney) Prove Theorem 20.4: The uniform topology on $\mathbb{R}^J$ is finer than the product topology and coarser than the box topology; these three topologies are different if $J$ is infinite. \\ \\ \noindent \begin{bf}8.1A\end{bf} Show the uniform topology is finer than the product topology. \\ \\ \noindent Take a basis element in the product topology: $\Pi U_\alpha$, where each $U_\alpha$ is open in $\mathbb{R}$. Show $\Pi U_\alpha$ is open in the uniform topology. Take a point $x$ in $\mathbb{R}^J$; represent $x$ as $x = (x_\alpha)$ for $\alpha \in J$. By definition of the product topology, there only finitely many of the $U_\alpha \neq \mathbb{R}$, so denote let $\alpha_1, \alpha_2, ..., \alpha_n$ be the values of $\alpha$ for which $U_\alpha \neq \mathbb{R}$. Next, for each $i$, choose $\epsilon_i > 0$ so that $B_{\overline{\rho}}(x_{\alpha_i}, \epsilon_i) \subset U_{\alpha_i}$, which is possible since by definition each $U_{\alpha_i}$ is open in $\mathbb{R}$. Claim: if we let $\epsilon = \min \{ \epsilon_1, ..., \epsilon_n \}$, $x \in B_{\overline{\rho}}(x, \epsilon) \subset \Pi U_\alpha$. If a point $z \in B_{\overline{\rho}}(x, \epsilon) then $\overline{d}(x_\alpha,z_\alpha) < \overline{\rho}(x,z) < \epsilon \leq \epsilon_i$ for all $i$ and $\alpha$. Hence $z_{\alpha_i} \in B_{\overline{\rho}}(x_{\alpha_i}, \epsilon_i) \subset U_{\alpha_i}$ for all $\alpha_i$ and $z_\alpha \in R = U_\alpha$ for all $\alpha \not= \alpha_1, ..., \alpha_n$, so $z \in \Pi U_\alpha$. Hence the uniform topology is finer than the product topology. 8.1B For $J$ infinite, show that the uniform topology is different than the product topology. $B_{\overline{\rho}}({\bf 0}, 1) = \Pi_{\alpha \in J} (-1, 1)$ which is not open in the product topology when $J$ is infinite. Thus, the uniform topology is different than the product topology when $J$ is infinite. 8.1C Show that the box topology is finer than the uniform topology. Note the book's proof is incorrect. Begin by letting $B$ be the $\epsilon_1$-ball around $y$ in the uniform topology. Show $B$ is open in the box topology. Take $x \in B$. Then there exists a $B_{\overline{\rho}}(x, \epsilon) \subset B_{\overline{\rho}}(y, \epsilon_1)$ Then we could define a neighbourhood of $x$ in the box topology as \[ U = \prod (x_\alpha - \frac{1}{2}\epsilon, x_\alpha + \frac{1}{2}\epsilon), \]$. Claim $U$ is contained in $B$. For any element $y$ of $U$, $\overline{d}(x_\alpha,y_\alpha) < \frac{1}{2}\epsilon$ for all $\alpha$. This implies that $\overline{\rho}(x,y) \leq \frac{1}{2}\epsilon$, implying that the box topology is indeed finer than the uniform topology. 8.1D Show that for $J$ infinite, the uniform topology and the box topology are different. Take $I \subset J$ where $I$ is a countable subset of $J$. Let $x_{n_\alpha} = \cases { {1 \over n} & $\alpha \in I$ \cr 0 & $\alpha \not\in I$}$. If ${\bf x}_n = \Pi{\alpha \in J} x_\alpha$ and ${\bf x} = \Pi_{\alpha \in J} 0$, then $\overline{\rho({\bf x_n}, {\bf 0}) = {1 \over n}$. Hence ${\bf x_n} \rightarrow {\bf 0}$ in the uniform topology, but ${\bf x_n}$ does not converge to ${\bf 0}$ in the box topology. Thus, the uniform topology and the box topology are different when $J$ is infinite. 8.2: Using the metric $D$ defined in Theorem 20.5, describe $B_D (0, \frac{1}{3})$. First off we note that $D(0,x) < \frac{1}{3}$ for $x$ to be an element of $B_D$. But \[D(0,x) = sup \{ \frac{ min \{ d(0,x_i),1 \} }{i} \} = \sup \{ \frac { min \{|x_i|,1 \} }{i} \} < \frac{1}{3} \]$ for each $i$. For $i = 1$, any point $x$ where $|x_1| < \frac{1}{3}$ will satisfy the above equation. For $i=2$, any point $x$ where $|x_2| < \frac{2}{3}$ will work. For $i=3$, any point $x$ where $|x_3| < 1$ will work. For $4 \leq i$, any point $x$ where $x_4 \in \mathbb{R}$ will work. Hence \[ B_D(0,\frac{1}{3}) = (-\frac{1}{3}, \frac{1}{3}) \times (-\frac{2}{3}, \frac{2}{3}) \times (-1, 1) \times \mathbb{R} \times \mathbb{R} \times ... \] p. 145: Repeat 4A for the equivalence relation $x_0 \times y_0 ~ x_1 \times y_1$ if $x_0^2 + y_0^2 = x_1^2 + y_1^2$. $f: X* \rightarrow [0, \infty), ~f((x, y) ) = \sqrt(x^2 + y^2)$ is a homeomorphism where $[0, \infty)$ is a subspace of $R$. Thus $X*$ is equivalent to the subspace $[0, \infty)$ of $R$. pg 152 1.) Let ${\cal T'}$ and ${\cal T}$ be two topologies on $X$, where ${\cal T'$ is finer than ${\cal T}$. What does the connectedness of $X$ in one imply about its connectedness in the other? $R_l$ is finer than $R$ which is finer than $R$ with the indiscrete topology, $R$ and $R$ with the indiscrete topology are connected but $R_l$ is not connected by problem 7. Thus connectedness of $X$ in ${\cal T}$ does not imply anything about the connectedness of $X$ in ${\cal T}'$ Suppose ${\cal T}$ is not connected, then $X = U \cup V$ where $U, V \in {\cal T} \subset {\cal T}'$. Thus, ${\cal T}'$ is not connected. Hence connectedness of $X$ in ${\cal T}'$ implies connectedness of $X$ in ${\cal T}$. 7.) $$R_l$ is not connected since $R_l = (\infty, 0) \cup [0, \infty)$ and $(\infty, 0)$ and $[0, \infty)$ are both open in $R_l$. p. 157 8A: If two spaces $X$ and $Y$ are path connected, is the space $X \times Y$ path connected? Let $a \times b$ and $c \times d$ be distinct points in $X \times Y$. Since $X$ is path connected, there exists a path $\gamma_1: [0, 1] \rightarrow X$ from $a$ to $c$. Similarly, since $Y$ is path connected, there exists a path $\gamma_2 [0, 1] \rightarrow Y$ from $b$ to $d$, Hence $\gamma = (\gamma_1, \gamma_2): [0, 1] \rightarrow X \times Y$ is a path from $a \times b$ to $c \times d$, so $X \times Y$ is indeed path connected. 8B, If $A \subset X$ and $A$ is path connected, is $\overline{A}$ also path connected? Not necessarily. Take $X = \mathbb{R}^2$ and $A = S$, where $S$ is the topologist's sine curve given in example 7, pg 157. $S$ is path connected, but $\overline{S}$ is not. 8CIf $f: X \rightarrow Y$ is continuous and $X$ is path connected, is $f(X)$ necessarily path connected? Fix $x,y$ in $f(X)$. Take $c \in f^{-1}(x)$ and $d \in f^{-1}(y)$, the inverse images of $x$ and $y$. Then since $X$ is path connected, there exists a continuous function $\gamma_X: [a,b] \mapsto X$, where $[a,b]$ is a subset of $\mathbb{R}$ such that $\gamma_X(a) = c$ and $\gamma_X(b) = d$. Since both $f$ and $\gamma_X$ are continuous, the composition of $f$ with $\gamma_X$ will also be a continuous function. Let this function be $\gamma$. Thus $\gamma = f \circ \gamma_X: [a,b] \mapsto f(X)$, where $\gamma(a) = f \circ \gamma_X(a) = f(c) = x$ and $\gamma (b) = f \circ \gamma_X(b) = f(d) = y$, so $f(X)$ is path connected. 8D If $\{ A_\alpha \}$ is a collection of path-connected subspaces of $X$, and if $\cap A_\alpha \neq \emptyset$, is $\cup A_\alpha$ necessarily path connected? \\ \\ \noindent Since $\cap A_\alpha \neq \emptyset$, by the definition of intersection there exists a point $y$ such that $y \in A_\alpha$ for all $\alpha$. Let $x, z$ be two points in $\cup A_\alpha$. $x$ therefore must be an element of $A_\alpha$ for some $\alpha$ by the definition of union; similarly for $z$. Let $x \in A_1$ and $z \in A_2$. Since $A_1$ is path connected, and $x,y \in A_1$, there exists a path $\gamma_1: [0, 1] \rightarrow A_1$ between $x$ and $y$. Similarly, there exists a path $\gamma_2: [1, 2] \rightarrow A_2$ between $y$ and $z$ in $A_2$. Hence $\gamma[0, 2] \rightarrow \cup A_\alpha, \gamma(x) = \cases{\gamma_1(x) & $x \in [0, 1] \cr \gamma_2(x) & $x \in [1, 2] $ is a path connecting $x$ and $z$ via $y$, a point in common. p. 170 1a.) (R, indiscrete topology) is compact, but (R, discrete topology) is not compact. Thus ${\cal T} \subset {\cal T}'$, $X$ compact w.r.t ${\cal T}$ does not imply $X$ is compact w.r.t ${\cal T}$. Suppose ${\cal T} \subset {\cal T}'$ and $X$ is compact w.r.t ${\cal T}'$. Let $X \subset \cup_{a \in A} U_a$, where $U_a \in {\cal T} \subset {\cal T}'$. Since $X$ is compact w.r.t ${\cal T}'$ and $U_a \in {\cal T}'$, there exists $a_1, ..., a_n \in A$ such that $X \subset \cup_{i = 1}^n U_{a_i}$. Thus, $X$ is compact w.r.t ${\cal T}$ 1b.) Suppose that $X$ is compact Hausdorff under both ${\cal T}$ and ${\cal T}'$ and that ${\cal T}$ and ${\cal T}'$ are comparable. Since ${\cal T}$ and ${\cal T}'$ are comparable, then either ${\cal T} \subset {\cal T}'$ or ${\cal T}' \subset {\cal T}$. Without loss of generality, assume ${\cal T} \subset {\cal T}'$. Suppose $A$ is a closed set in ${\cal T}'$. Then $A$ is compact w.r.t ${\cal T}'$. Hence by 1a, $A$ is compact w.r.t ${\cal T}$. Hence $A$ is closed in ${\cal T}$. Thus, ${\cal T} = {\cal T}'$. 2a.) Let $A \subset R$ where $R$ has finite complement topology. Let $A \subset \cup_{d \in D} U_d$ be an open cover of $A$. Take $d_0 \in D$ such that $U_{d_0} \not= \emptyset$. Since $R - U_{d_0}$ is finite, $A- U_{d_0}$ is also finite and hence $A- U_{d_0}= \{a_1, ..., a_n\}. For each $i = 1, ..., n$, take $d_i \in D$ such that $x_i \in U_{d_i}$. Then $A \subset \cup_{i = 0}^n U_{d_i}$. Hence $A$ is compact. 2b.) Let $U_n = R - \{ n, n + 1, n + 2, ...\}$. Then $R \subset \cup_{i = 1}^\infty U_{n}$, but there does not exist a finite subcover. Hence $R$ with the co-countable topology is not compact. 4.) Suppose $A \subset X$ is not bounded where $X$ is metrizable. Take $a \in A$. Then $A \subset \cup_{n = 1}^\infty B(a, n)$, but there does not exist a finite subcover of this cover (for otherwise $A$ would be bounded). Hence $A$ is not compact. Suppose $A$ is not closed. Then there exists $x_0 \in A' - A$. Since $x_0$ is a limit point of $A$. There exists $a_n \in B(x_0, {1 \over n}) \cap A - x_0$. Note that $U = \{x \in X ~|~ d(x, x_0) > {1 \over 4}\}$ is open in $X$. Thus $A \subset \cup_{n = 1}^\infty B(a_n, {1 \over 2n}) \cup U$, but there does not exist a finite subcover of this cover. Hence $A$ is not compact. $R$ with the discrete metric is closed and bounded, but not compact. 5.) Since $A$ and $B$ are closed, you can use thm 32.4 p. 126 1.) 2.) $d((x_1, y_1),