%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% exercise set 5 - solutions %%% October 9, 2009 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass{amsart} \usepackage{graphicx} \usepackage{amssymb} \usepackage{epstopdf} \usepackage{nicefrac} \DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png} % \input hw.sty \textwidth = 6 in \textheight = 8.5 in \oddsidemargin = 0.0 in \evensidemargin = 0.0 in \topmargin = 0.0 in \headheight = 0.0 in \headsep = 0.0 in \parskip = 6pt \parindent = 0.0in % \hoffset=-.7in % \voffset=-.7in \newcommand{\mN}{{\mathbb N}} \newcommand{\mR}{{\mathbb R}} \newcommand{\mZ}{{\mathbb Z}} \begin{document} \begin{center} Math 445, Fall 2009 \hfill Exercise Set \#5 \hfill Solutions \end{center} \bigskip {\bf 1.} [\#10, page 78] ~ Let $A, B \subset M$ be subsets of a metric space $M$. Define the distance between the sets to be $$D(A,B) = \inf\{ D(a,b) \mid a \in A ~ , ~ b \in B\}$$ a) Suppose that $B = \{x\}$ consists of a single point. Prove that $D(A,B) = 0$ if and only if $x \in \overline{A}$. \medskip b) Give an example in the Euclidean plane of two closed subsets, $A, B \subset \mR^2$, such that $A \cap B = \emptyset$ and yet $D(A,B) = 0$. [Hint: the sets $A$ and $B$ cannot be bounded.] \proof a) Suppose that $x \in \overline{A}$. For every $\epsilon > 0$ the intersection $B(x,\epsilon) \cap A \ne \emptyset$ so there exists $ y \in B(x,\epsilon) \cap A$ with $D(x,y) < \epsilon$. Thus, $D(A, \{x\}) < \epsilon$ for all $\epsilon > 0$, so $D(A, \{x\}) = 0$. Conversely, suppose that $D(A, \{x\}) = 0$. Then by the definition of the infimum, for every $\epsilon > 0$ there exists some $y \in A$ with $D(y,x) < \epsilon$. This implies $B(x,\epsilon) \cap A \ne \emptyset$ for all $\epsilon > 0$, hence for every open set $U$ with $x \in U$ we have $A \cap U \ne \emptyset$, hence $x \in \overline{A}$. b) There are lots of ways this can happen; here is one: Let $A = \{(x,0) \mid x \in \mR\} \subset \mR^2$ be the $x$-axis. Let $B = \{(x, e^x) \mid x \in \mR\}$ be the graph of the exponential function. Both sets are closed in $\mR^2$ as their complements are open. [obvious?] For each integer $n \geq 1$, set $a_n = (-n, 0) \in A$ and $b_n = (-n, e^{-n}) \in B$. Then $D(a_n, b_n) = e^{-n} \to 0$, so $D(A,B) = 0$. \endproof \medskip {\bf 2.} [\#2, page 82] ~ Let $u \in M$ be a point in a metric space $M$. The function $f(x) = D(u,x)$ maps $M$ into the real numbers, $f \colon M \to \mR$. Prove that $f$ is continuous. \proof We use the $\epsilon-\delta$ criteria for continuity. For each $x_0 \in M$ then we show $f(x) = D(u,x)$ is continuous at $x_0$. Let $\epsilon > 0$ be given. Translating the condition for continuity into this case, we need to find $\delta > 0$ such that, if $D(x_0, x) < \delta$ then the distance from $u$ to $x$ is within $\epsilon$ of the distance from $u$ to $x_0$. The claim is that we can take $\delta = \epsilon$ then this follows from the triangle inequality. [A picture might help.] Assume that $D(x_0, x) < \epsilon$ then $$f(x) = D(u,x) \leq D(u,x_0) + D(x_0 , x) < f(x_0) + \epsilon$$ so $f(x) - f(x_0) < \epsilon$. Similarly, $$f(x_0) = D(u,x_0) \leq D(u,x) + D(x , x_0) < f(x) + \epsilon$$ so $f(x_0) - f(x) < \epsilon$, hence we have $| f(x) - f(x_0) | < \epsilon$. \endproof \medskip {\bf 3.} [\#3, page 82] ~ Let $A \subset M$ be a fixed subset of a metric space $M$. The function $f(x) = D(A,x)$ maps $M$ into the real numbers, $f \colon M \to \mR$. Prove that $f$ is continuous. \proof Again, we use the $\epsilon-\delta$ criteria for continuity. The difference from the case in Problem 2 above is that the distance to the set $A$ is the infimum over the distances to all points of $A$. The idea is to pick points $u \in A$ close to $x_0$ and $v \in A$ close to $x$, and make the estimates as above, but with some ``room for error''. This will be provided by using a ``two epsilon'' trick. Let $x_0 \in M$ then we show $f(x) = D(A,x)$ is continuous at $x_0$. Let $\epsilon > 0$ be given, then take $\delta = \epsilon/2$. Let $D(x_0, x) < \epsilon/2$ and choose $u \in A$ so that $D(u,x_0) < D(A,x_0) + \epsilon/2$. \begin{eqnarray*} f(x) = D(A,x) & = & \inf \{D(y,x) \mid y \in A\} \\ & \leq & D(u,x)\\ & \leq & D(u,x_0) + D(x_0, x) \\ & < & D(A,x_0) + \epsilon/2 + \epsilon/2\\ & = & f(x_0) + \epsilon \end{eqnarray*} so $f(x) - f(x_0) < \epsilon$. Similarly, choose $v \in A$ so that $D(v,x) < D(A,x) + \epsilon/2$. Then \begin{eqnarray*} f(x_0) = D(A,x_0) & = & \inf \{D(y,x_0) \mid y \in A\} \\ & \leq & D(v,x_0)\\ & \leq & D(v,x) + D(x, x_0) \\ & < & D(A,x) + \epsilon/2 + \epsilon/2\\ & = & f(x) + \epsilon \end{eqnarray*} so $f(x_0) - f(x) < \epsilon$, hence we have $| f(x) - f(x_0) | < \epsilon$. \endproof \medskip {\bf 4.} [\#4, page 82] ~ Let $A \subset M$ be a \emph{closed} subset and $y$ a point in a metric space $M$, with $y \not\in A$. Prove that there exists a continuous real-valued function on $M$ which vanishes on $A$ but not at $y$. \proof Since $A$ is closed, the complement $M-A$ is open, so there exists $\epsilon > 0$ so that the open ball $B(y, \epsilon) \subset M- A$. But this means that for all $u \in A$, the distance $D(u, y) \geq \epsilon$. Hence, $D(A, y) > 0$. Define $f(x) = D(A,x)$, which is continuous by Problem 3, and $f(y) \geq \epsilon > 0$ by the above. The property $f(u) = 0$ for $u \in A$ is obvious. \endproof \medskip {\bf 5.} ~ Let $A \subset M$ be a subset and $y$ a point in a metric space $M$. Suppose that $y \not\in \overline{A}$. Prove that there exists a continuous real-valued function $f \colon M \to [0,\infty)$ which vanishes on $A$ but not at $y$. \proof Since $y \in M - \overline{A}$, we can use Problem 4 for the closed set $\overline{A}$ to define a continuous function $f(x) = D(\overline{A}, x)$ which vanishes on $\overline{A}$, so also vanishes on $A$, but $f(y) > 0$. \endproof \vfill \end{document}