\magnification 1200 \parindent 0pt \parskip 10pt \hsize 7.2truein \vsize 9.7 truein \hoffset -0.45truein \def\u{\vskip -7pt} \def\v{\vskip -6pt} p91: 4, 6 4.) Let $W$ be open in $X \times Y$. Show $\pi_1(W)$ is open in $X$. Let $x_0 \in \pi_1(W)$. Find $U$ open in $X$ such that $x_0 \in U \subset \pi_1(W)$. $x_0 \in \pi_1(W)$ implies there exists an $(x_0, y_0) \in W$ such that $\pi_1(x_0, y_0) = x_0$. $W$ open implies there exists a basis element $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$ and $(x_0, y_0) \in U \times V \subset W$ Thus $\pi_1((x_0, y_0)) \in \pi_1(U \times V) \subset \pi_1(W)$. Hence $x_0 \in U \subset \pi_1(W)$. 6.) $R^2$ has the product topology. Thus by thm 15.1, $ \{B_1 \times B_2 ~|~ B_1 \in {\cal B}_1, B_2 \in {\cal B}_2 \} $ is basis for $R^2$ where ${\cal B}_i$ are bases for $R$. By problem \#8, p. 83, ${\cal B} = \{(a, b) ~|~ a < b, ~a$ and $b$ rational $\}$ is a basis for $R$. Thus $\{(a, b) \times (c, d) ~|~ a < b$ and $c < d, ~a, b, c, d$ rational $\}$ is a basis for $R^2$. p91: 1, 3, 8 1.) Let ${\cal T}_X$ be the topology on $X$. Let ${\cal T}_Y = \{ Y \cap U ~|~ U \in {\cal T}_X\}$ Let ${\cal T}_A = \{ A \cap U ~|~ U \in {\cal T}_X\}$ Let ${\cal T}_A' = \{ A \cap U ~|~ U \in {\cal T}_Y\}$ Show ${\cal T}_A = {\cal T}_A'$ Let $W \in {\cal T}_A$. Then there exists a $U \in {\cal T}_X$ such that $W = A \cap U$. $U \in {\cal T}_X$ implies $Y \cap U \in {\cal T}_Y$. Since $A \subset Y$, $A \cap (Y \cap U) = (A \cap Y) \cap U = A \cap U = W$. Thus $W \in {\cal T}_A'$. Hence ${\cal T}_A \subset {\cal T}_A'$ Let $W \in {\cal T}_A'$. Then there exists a $U \in {\cal T}_Y$ such that $W = A \cap U$. $U \in {\cal T}_Y$ implies there exists a $V \in {\cal T}_X$ such that $U = Y \cap V$. Since $A \subset Y$, $W = A \cap U = A \cap Y \cap V = A \cap V$. Thus $W \in {\cal T}_A$. Hence ${\cal T}_A' \subset {\cal T}_A$ Therefore ${\cal T}_A = {\cal T}_A'$ 3.) A and E are open in $R$. $B, C, D$ are not open in $R$. $A, B, E$ are open in $[-1, 1]$. $C, D$ are not open in $[-1, 1]$. $A = (-1, -{1 \over 2}) \cup ({1 \over 2}, 1)$ is open in $R$ since $A$ is the union of two basis element. $B = [-1, -{1 \over 2}) \cup ({1 \over 2}, 1]$. Show $B$ is not open in $R$. Show that there exists an $x \in B$ such that for all basis elements, $B_x$ such that $x \in B_x$, $B_x \not\subset B$. Let $x = 1$. Suppose $1 \in (a, b)$. Thus $a < 1 < b$. Hence $a < 1 < { 1 + b \over 2} < b$. Thus ${ 1 + b \over 2} \in (a, b)$, but ${ 1 + b \over 2} \not\in [-1, -{1 \over 2}) \cup ({1 \over 2}, 1]$. Thus, $(a, b) \not\subset B$. Thus $B$ is not open. Similarly $C$ (take $x = {1 \over 2}$) and $D$ (take $x = {1 \over 2}$ or 1) are not open. Take $x \in E =\{x ~|~ 0 < |x| < 1$ and ${1 \over x} \not\in Z_+\}$. If $x < 0$, then $x \in (-1, 0) \subset E$. If $x \not< 0$. Then $x \not = {1 \over n}$ for any $n \in Z_+$ and $x \not= 0$ since $x \in E$. Let $F = \{ n \in Z_+ ~|~ {1 \over n} < x\}$. Since $F$ is bounded below by 0, $inf ~F$ exists in $Z$ since $Z$ has the greatest lower bound property. Let $N = inf ~F$. Then ${1 \over N} < x$. $N-1 < N$ implies that $N-1 \not\in F$. Since $x < 1$, $N \not= 1$, and thus $N-1 \in Z_+$. Hence ${1 \over N-1} \geq x$ and since $x \not= {1 \over N-1}, ~{1 \over N-1} > x$. Thus, $x \in ({1 \over N}, {1 \over N-1}) \subset E$. Thus $E$ is open in $R$. Alternate proof that $E$ is open in $R$. $E = (-1, 0) \cup [ \cup_{n=1}^\infty ({1 \over n+1}, {1 \over n})]$. Thus $E$ is open since it is the union of open sets. Question: Can you prove that $E = (-1, 0) \cup [ \cup_{n=1}^\infty ({1 \over n+1}, {1 \over n})]$. $A$ open in $R$ and $A = A \cap [-1, 1]$ implies $A$ is open in $[-1, 1]$. $(-3, -{1 \over 2}) \cup ({1 \over 2}, 5)$ open in $R$ and $B = [(-3, -{1 \over 2}) \cup ({1 \over 2}, 5)] \cap [-1. 1]$ implies $B$ is open in $[-1. 1]$. $C = (-1, -{1 \over 2}] \cup [{1 \over 2}, 1)$ Show $C$ is not open. Show that there exists an $x \in C$ such that for all basis elements, $B_x$ such that $x \in B_x$, $B_x \not\subset C$. Let $x = {1 \over 2}$. Suppose ${1 \over 2} \in (a, b) \cap [-1, 1]$. Let $a' = max\{a, -{1 \over 2}\}$. Thus $a \leq a' < {1 \over 2} < b$. Hence $a < { {1 \over 2} + a' \over 2} < {1 \over 2} < b$. Also $-{1 \over 2} \leq a' < {1 \over 2}$ implies $-1 < -{1 \over 2} < { {1 \over 2} + a' \over 2} < {1 \over 2} < 1$ Thus ${ {1 \over 2} + a' \over 2} \in (a, b) \cap [-1, 1]$, but ${ {1 \over 2} + a' \over 2} \not\in (-1, -{1 \over 2}] \cup [{1 \over 2}, 1)$. Thus, $(a, b) \cap [-1, 1] \not\subset C$. Thus $C$ is not open. Similarly $D$ is not open. $E$ open in $R$ and $E = E \cap [-1, 1]$ implies $E$ is open in $[-1, 1]$ 8.) Let $R^2$ have the topology generated by ${\cal T}_X \times {\cal T}_Y$. Let $L$ be a line with slope 0. Then $L = R \times \{y_0\}$ for some $y_0 \in R$. Note that $L = R \times \{y_0\}$ has the subspace topology by hypothesis. By thm 16.3, the subspace topology on $L= R \times \{y_0\}$ as a subspace of $R^2$ is the same as the product topology on $R \times \{y_0\}$ where $R$ is a subspace of $(R, {\cal T}_X)$ and $\{y_0\}$ is a subspace of $(R, {\cal T}_Y)$. Thus, $R \times \{y_0\}$ has the topology generated by $\{ U \times V ~|~ U \in {\cal T}_X, V \in \{\emptyset, \{y_0\}\} \}$. Since $A \times \emptyset = \emptyset$, $\{ U \times V ~|~ U \in {\cal T}_X, V \in \{\emptyset, \{y_0\}\} \}$ = $\{ U \times V ~|~ U \in {\cal T}_X, V = \{y_0 \} \}$. Thus the topology on $L$ is equivalent to ${\cal T}_X$. Thus $L$ inherits the lower limit topology as a subspace of $R_{ l} \times R$ and $L$ inherits the lower limit topology as a subspace of $R_{ l} \times R_{ l}$. Similarly if $L$ is a line with vertical slope, the the topology on $L = {\cal T}_Y$. Thus $L$ inherits the standard topology as a subspace of $R_{ l} \times R$ and $L$ inherits the lower limit topology as a subspace of $R_{ l} \times R_{ l}$. Let $R^2$ have the topology generated by $R \times R_{ l}$. Let $L$ be a line with positive slope. Let $f(x) = mx + b$ where $m > 0$. $L = \{(x, f(x)) ~|~ x \in R \}$. $(a, b) \times [c, d)$ is a basis element for the topology generated by $R \times R_{ l}$. If $f(a) \geq d$ or $f(b) \leq c$, then $\{(x, f(x)) ~|~ x \in R\} \cap (a, b) \times [c, d) = \emptyset$ If $f(a) < c$ and $f(b) \geq d$ then $\{(x, f(x)) ~|~ x \in R \} \cap (a, b) \times [c, d) = [ (f^{-1}(c), c), (f^{-1}(d), d))$ where $[(f^{-1}(c), c), (f^{-1}(d), d)) = \{(x, f(x)) ~|~ f^{-1}(c) \leq x < f^{-1}(d)\}$ If $f(a) < c$, $c < f(b) \leq d$ then $\{(x, f(x)) ~|~ x \in R \} \cap (a, b) \times [c, d) = [(f^{-1}(c), c), (b, f(b)))$ where $[(f^{-1}(c), c), (b, f(b))) = \{(x, f(x)) ~|~ f^{-1}(c) \leq x < b, \}$ Thus since $f$ is onto, every half open interval in $L$ is open. Thus the topology on $L$ is equal to or finer that the lower limit topology. If $c \leq f(a) < d$ and $f(b) \geq d$ then $\{(x, f(x)) ~|~ x \in R \} \cap (a, b) \times [c, d) = ((a, f(a)), (f^{-1}(d), d))$ where $((a, f(a)), (f^{-1}(d), d)) = \{(x, f(x)) ~|~ a < x < f^{-1}(d)\}$ If $c \leq f(a) < d$, $c < f(b) \leq d$ then $\{(x, f(x)) ~|~ x \in R \} \cap (a, b) \times [c, d) = ((a, f(a)), (b, f(b))$ where $((a, f(a)), (b, f(b)) = \{(x, f(x)) ~|~ a < x < b, \}$ Thus, every open interval in $L$ is open. Note that open intervals are open in the lower limit topology. Since we have determined for every possible intersection between $L$ and basis elements of $R \times R_{ l}$, the collection of half open intervals is a basis for the topology on $L$. Thus $L$ inherits the lower limit topology as a subspace of $R \times R_{ l}$ Similarly, $L$ inherits the lower limit topology as a subspace of $R_{l} \times R_{ l}$ Similarly, if $L$ is a line with negative slope, $L$ inherits the upper limit topology as a subspace of $R \times R_{ l}$ Let $R^2$ have the topology generated by $R_{l} \times R_{ l}$ Let $L$ be a line with negative slope. $L = \{(x, y) ~|~ y = mx + b\}$ where $m < 0$. Let $f(x) = mx + b$. $[x, x+1) \times [f(x), f(x)+1) \cap L = \{(x, f(x))\}$. Thus every point in $L$ is open. Thus the topology on $L$ is discrete topology. \end