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\pset{5}{Dan Ports}{2003/04/11}{drkp@mit.edu}

\begin{ppart}{Problem 1}
\paragraph{Part a}
test
Since $\sum a_n$ diverges, there are two possibilities: either $\lim_{x \to \infty} a_n = 0$ or $\lim_{x \to \infty} \neq 0$. In the latter case, $a_n \to p$, for some extended real number $p \neq 0$. Then $\frac{a_n}{1 + a_n} \to \frac{p}{1 + p}$, which is clearly non-zero. So by the divergence test, $\sum \frac{a_n}{1 + a_n}$ diverges. Otherwise, $a_n \to 0$. In this case, we can thus find a $N > 0$ such that $\forall n \ge N$, $a_n < 1$ by the definition of convergence. Then $2(\frac{a_n}{1 + a_n}) < 2 (\frac{a_n}{2}) = a_n$. Thus, by the comparison test, $\sum 2(\frac{a_n}{1 + a_n})$ diverges since $\sum a_n$ diverges. Hence, $\sum \frac{a_n}{1 + a_n}$ diverges. Thus $\sum \frac{a_n}{1 + a_n}$ diverges for any $a_n$ where $\sum a_n$ diverges.

\paragraph{Part b}
We know $\forall n a_n > 0 $, so $s_{n+1} > s_n$ ($s_n$ is monotonically increasing. So:
\[\frac{a_{N+1}}{s_{N+1}} + ... + \frac{a_{N+k}}{s_{N+k}} \ge \frac{a_{N+1} + ... a_{N+k}}{s_{N+k}} = \frac{s_{N+k} - s_N}{s_{N+k}} = 1 - \frac{s_N}{s_{N+k}}\]
We can use this to show that $\sum \frac{a_n}{s_n}$ is divergent. Let $S_n$ be the $n$th partial sum of this sequence: $S_n = \sum_{i=1}{n}\frac{a_i}{s_i}$. Then we consider the difference between two partial sums $S_{N+k}$ and $S_{N}$:
\[S_{N+k}-S_{N} = \frac{a_{N+1}}{s_{N+1}} + ... + \frac{a_{N+k}}{s_{N+k}} \ge 1 - \frac{s_N}{s_{N+k}}\]
Since $\sum a_n$ diverges, $\lim_{n \to \infty} s_n = \infty$, and if we fix $N$, $\lim_{k \to \infty} \frac{s_N}{s_{N+k}} = 0$. So $\lim_{k \to \infty} 1 - \frac{s_N}{s_{N+k}} = 1$. Combined with the inequality above, this means that, for any $N$ we can find a $k$ such that the difference between the $N+k$th and $N$th partial sums is at least $1 - \epsilon$, for any $\epsilon > 0$. Arbitrarily choosing $\epsilon = \frac{1}{2}$, we can find a $k$ for any $N$ such that $S_{N+k} - S_N \ge \frac{1}{2}$. This means that the sequence of partial sums $S_n$ does not satisfy the Cauchy criterion, so it does not converge, and thus $\sum \frac{a_n}{s_n}$ diverges.

\paragraph{Part c}
\begin{eqnarray*}
\frac{a_n}{s_n^2} &=& \frac{a_n}{s_n s_n} \\
& \le & \frac{a_n}{s_n s_{n-1}}
 \textrm{ by the monotonicity of $s_n$ as shown above} \\
 &=& \frac{s_n - s_{n-1}}{s_n s_{n-1}} \textrm { by definition of $s_i$} \\
 &=& \frac {s_n}{s_n s_{n-1}} - \frac{s_{n-1}}{s_n s_{n-1}} = \frac {1}{s_{n-1}} - \frac{1}{s_n} \\
\end{eqnarray*}
\paragraph{}
We use this to show that $\sum \frac{a_n}{s_n^2}$ converges.  Using the inequality above, this is less than the series $\sum_{n=2}^\infty \frac {1}{s_{n-1}} - \frac{1}{s_n}$. Consider the partial sums $S_N = \sum_{n=2}^N \frac {1}{s_{n-1}} - \frac{1}{s_n} = \frac {1}{s_1} - \frac{1}{s_2} + \frac {1}{s_2} - \frac{1}{s_3} + ... + \frac {1}{s_{N-1}} - \frac{1}{s_N}$. This is a telescoping sum which cancels to $\frac {1}{s_1} - \frac{1}{s_N} = \frac {1}{a_1} - \frac{1}{s_N}$. We know $\frac {1}{a_1}$ is constant, and $s_n \to \infty$ because $a_n$ diverges, so $\lim_{N \to \infty} S_N = \frac {1}{a_1}$. Thus the sequence of partial sums converges, so we have shown that $\sum \frac{a_n}{s_n^2}$ is less than a series that converges, so it converges by the comparison test.

\paragraph {Part d}
Nothing can be said about the convergence of the first series. We will show an example of an appropriate series $a_n$ that makes $\sum \frac{a_n}{1+n a_n}$ diverge, and another choice for $a_n$ that makes it converge. First, consider $\sum \frac{a_n}{1+n a_n}$. If $a_n = 1$ (clearly $\sum a_n$ is divergent), the series becomes $\sum \frac{1}{1+n}$. Each term is larger than $\frac{1}{2n}$ (for $n>2$), which diverges (since it is a multiple of $\frac{1}{n}$), so by the comparison test $\sum \frac{a_n}{1+n a_n}$ diverges. 
\paragraph{}
Next consider a sequence $a_n$ that equals $1$ when n is a perfect square and $\frac{1}{n^2}$ otherwise. Then $\sum a_n$ diverges because its subseries $\sum_{\{n \textrm{ a perfect square}\}} 1$ certainly diverges. We will show, however, that $\sum \frac{a_n}{1+n a_n}$ converges. We can write this series as $\sum b_n + \sum_{n \textrm{a perfect square}}\frac{1}{1+n}$, where $b_n = \frac{\frac{1}{n^2}}{1+n\frac{1}{n^2}} = \frac{1}{1+n^2}$ when $n$ is not a perfect square and $0$ when it is. By the comparison test, $\sum b_n$ converges because $\forall n~ b_n \leq \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges. And we can write $\sum_{\{n \textrm{ a perfect square}\}}\frac{1}{1+n}$ as $\sum_{m=1 }^{\infty}\frac{1}{1+m^2}$ (letting $n = m^2$). This series converges, since every term is less than $1/m^2$, which is a convergent series. Thus $\sum \frac{a_n}{1+n a_n}$ is a sum of two convergent series and converges itself for this choice of $a_n$.
\paragraph{}
We show that $\sum \frac{a_n}{1+n^2 a_n}$ always converges. Note that $\frac{a_n}{1+n^2 a_n} = \frac{1}{\frac{1}{a_n}+n^2}$. Since $\forall n~a_n>0$ by definition, the denominator is always greater than $n^2$, and so the terms of the series are less than those of the series $\frac{1}{n^2}$, which we know converges. So by the comparison test, $\sum \frac{a_n}{1+n^2 a_n}$ always converges for any such $a_n$.

\end{ppart}

\begin{ppart}{Problem 2}
\paragraph{}
Let $f$ be a continuous function from a metric space $X$ to a metric space $Y$, and $E$ be a subset of $X$. For any $p \in E$, clearly $f(p) \in f(E) \subseteq \overline{f(E)}$. Now consider $p \in E'$. Then $p$ is a limit point of $E$. We will show that $f(p)$ is a limit point of $f(E)$. That is, given any $\epsilon > 0$, we will find a point $q \in E$ such that $d_Y(f(p),f(q)) < \epsilon$. Since $f$ is continuous, there exists a $\delta$ such that $\forall x \in X$ where $d_X(p, x) < \delta$, $d_Y(f(p), f(x)) < \epsilon$. Since $p$ is a limit point of $E$, we can find a point $q \in E$ such that $d_X(p,q) < \delta$ and thus $d_Y(f(p), f(q)) < \epsilon$. So $f(p)$ is a limit point of $f(E)$. Thus, $\forall p \in E'$, $f(p) \in \overline{f(E)}$
\paragraph{}
We have shown that $f(E) \subseteq \overline{f(E)}$ and that $f(E') \subseteq \overline{f(E)}$. So since $\overline{E} = E \cup E'$, $f(\overline{E}) \subseteq \overline{f(E)}$.
\paragraph{}
As an example, consider $f$ from $R$ to $R$ such that $f(x) = \frac{1}{x}$. Let $E = {1,2,3,4...}$. Clearly, E has no limit points. But $f(e) = {\frac{1}{1},\frac{1}{2},\frac{1}{3},...}$, which has a limit point at zero. $0$ is in $\overline{f(E)}$ but not in $f(\overline{E})$, so $f(\overline{E}) \subset \overline{f(E)}$.
 \end{ppart}

\begin{ppart}{Problem 3}
\paragraph{}
Let $f$ and $g$ be continuous mappings from a metric space $X$ to a metric $Y$, and $E$ be a dense subset of $X$. This means that $\overline{E} = X$. From problem 2, $f(\overline{E}) \subseteq \overline{f(E)}$. Thus, $f(X) = f(\overline{E}) \supseteq \overline{f(E)}$. Thus any point in $f(X)$ is in the closure of $f(E)$, so it is either a point of $f(E)$ or a limit point of $f(E)$. By definition of density, $f(X)$ is dense in $f(E)$.
\paragraph{}
Now suppose that $\forall p \in E~g(p) = f(p)$. We will show by contradiction that $\forall p \in X~g(p) = f(p)$. Suppose there is some point $q \in X$ such that $g(q) \neq f(q)$ (it must be the case that $q \notin E$). Because $E$ is dense in $X$, $q$ is a limit point of $E$. So we can construct a sequence $\{p_1, p_2, p_3, ...\}$ of elements of $E$ such that the sequence $\{p_i\}$ converges to $q$. Because $f$ is continuous, the sequence $\{f(p_i)\}$ converges to $f(q)$ since $\{p_i\}$ converges to $q$. Because each $p_i$ is in $E$, $\forall i~f(p_i)=g(p_i)$; thus, $\{g(p_i)\}$ must also converge to $f(q)$ (two convergent sequences cannot converge to different points). But $f(q) \neq g(q)$, so $\{g(p_i)\}$ does not converge to $g(q)$, contradicting the continuity of $g$. Therefore, there can be no such point $q$ and $\forall p \in X~g(p) = f(p)$.
\end{ppart}

\begin{ppart}{Problem 4}
\paragraph{}
First, note that if $E$ is closed, then it is easy to show that $f(E)$ is bounded. In this case, $E$ is a closed and bounded subset of $R$, so $E$ is compact. $f(E)$ is uniformly continuous, so it is continuous, and it is defined on $E$, so $f(E)$ is therefore also compact. Compactness implies closed and boundedness, so $f(E)$ must be bounded.
\paragraph{}
We will now show that this is also true in other cases. Let $E$ be a bounded set in $R$. There is thus some number $M \in R$ such that $\forall x \in E ~ \abs{x} < M$. We first show that the closure of $E$ is also bounded. Consider a limit point $p$ of $E$. Because $p$ is a limit point, for every $\epsilon > 0$ there exists a point $q$ in $E$ such that $\abs{p-q} < \epsilon$. Using the triangle inequality between $p$, $q$, and $0$, $\abs{q} \leq \abs{p-q} + \abs{p} < \epsilon + M$. Since we can make $\epsilon$ arbitrarily small, for any $K > M$, $\forall x \in \overline{E} ~ \abs{x} < K$. So $\overline{E}$ is bounded. It is also closed by definition, and it is a subset of $R$, so it is compact.
\paragraph{}
Let $f$ be a uniformly continuous function from $E$ to $R$. By definition of uniform continuity, for every $\epsilon > 0$ we can find a $\delta > 0$ such that $\forall p \in E ~ \forall q \in E$ such that $\abs{p-q} < \delta$, $\abs{f(p)-f(q)} < \epsilon$. We arbitrarily choose 1 as a positive value for $\epsilon$ and use the uniform continuity of $f$ to obtain the corresponding value $\delta$ such that $\forall p \in E ~ \forall q \in E$ such that $\abs{p-q} < \delta$, $\abs{f(p)-f(q)} < 1$. Next we consider the neighborhoods of radius $\delta$ around every point $x \in E$: $N_\delta(x) = \{p \in E ~|~ \abs{x-p} < \delta\}$. Certainly every point of $E$ is in one of these neighborhoods. Also, every limit point $x$ of $E$ is contained in one of these neighborhoods: since $x$ is a limit point, we can find a point $q \in E$ such that $\abs{x-q} < \delta$, and so $x \in N_\delta(q)$. Since the neighborhoods are open sets, their union is an open set, and so because $\bigcup_{x\in E} N_\delta(x)$ contains $\overline{E}$, it is an open cover of $\overline{E}$. $\overline{E}$ is compact, so the open cover contains a finite subcover: a set of points $\{x_1, x_2, ..., x_n\}$ such that $\bigcup_{i = 1}^n N_\delta(x_i)$ contains $\overline{E}$. It clearly also contains $E$.
\paragraph{}
Since each point $x_i$ is in E, $f(x_i)$ is defined. Note that every point of E is within $\delta$ of one of the $x_i$'s, since $E$ is contained in their union of radius-$\delta$ neighborhoods. So because of our choice of $\delta$ based on uniform continuity, $\forall x \in E, \exists i$ such that $\abs{f(x)-f(x_i)} < 1$. Since there are a finite number of points $x_i$, we can find $\max\{\abs{f(x_i)}\}$. Now we observe that for every $x$ in $E$, $\abs{f(x)}$ cannot be less than (by the triangle inequality) the distance from $f(x)$ to the nearest $f(x_i)$, which we just showed was less than $1$, plus $\abs{f(x_i)}$, which cannot be greater than the maximum value $\max\{\abs{f(x_i)}\}$. So $\forall x \in E, \abs{f(x)} < 1 + \max\{\abs{x_i}\}$, which means $f(x)$ is bounded.
\paragraph{}
We give an example that shows that this does not apply if $E$ is unbounded. Consider the function $f(x) = \sqrt{x}$, defined on the interval of $[0,\infty)$. We will show that $f(x)$ is uniformly continuous. Given some $\epsilon > 0$, let $\delta = \epsilon^2$. Then, for any numbers $x$ and $y$ that satisfy $\abs{y-x} < \delta$, assume without loss of generality that $y > x$. Then $f(y)-f(x) = \sqrt{y} - \sqrt{x} < \sqrt{x+\delta} - \sqrt{y} \leq \sqrt{x} + \sqrt{\delta} - \sqrt{x} = \sqrt{\delta} = \epsilon$. So $f(x)$ is uniformly continuous on this domain. Also note that $f(x) = \sqrt{x}$ is obviously unbounded on this infinite interval.
\end{ppart}

\begin{ppart}{Problem 5}
\paragraph{}
We will first show that $\lim_{t \to x} f(t) = 0$ for every real number $x$. Let $\{p_1, p_2, p_3, ...\}$ be a sequence of real numbers that converges to $x$. We will show that $\{f(p_n)\}$ converges to $f(x)$. For every $n$ such that $p_n$ is irrational, $f(p_n) = 0$ by definition of the function. Therefore, for any subsequence $\{p_{q_n}\}$ of irrational numbers in $\{p_n\}$, the corresponding subsequence $\{f(p_{q_n})\}$ of $\{f(p_n)\}$ will converge to $0$. Therefore, we will have shown $\{f(p_n)\}$ converges to $0$ if we can verify that for every sequence of rational numbers $\{a_n\}$ that converges to $x$, the sequence $\{f(a_n)\}$ converges to $0$.
\paragraph{}
So we consider a sequence $\{a_n\}$ of rational numbers (write them as a ratio of two integers in lowest terms) that converges to $x$. We know such a convergent sequence exists because the rationals are dense in the reals. We will show that the denominators must approach infinity. Consider the set $\{b_n\}$ of rational numbers in the interval $(x-a_1,x+a_1)$ with (lowest-terms) denominator not greater than some arbitrary integer $k$. Because the denominator is not greater than $k$, the distance between any two elements in the sequence is not less than $\frac{1}{k}$. The interval has finite length (specifically, length $2a_1$), and the elements are spaced some minimum distance ($\frac{1}{k}$) apart, so there can only be a finite number of elements in the set (at most $2a_1 k$ elements). Thus, we can find a the minimum distance $\epsilon = \min\{\abs{x - b_n}$ such that no element in $\{b_n\}$ is within $\epsilon$ of $x$. We said earlier that $\{a_n\}$ was a sequence that converges to $x$, so we must be able to find an integer $N$ such that all elements in the sequence after $a_N$ are within $\epsilon$ of $x$. Since all elements in $\{b_n\}$ are at least $\epsilon$ away, $\forall m > N~a_m \notin \{b_n\}$. $\{b_n\}$ contains all elements closer to $x$ than $a_1$ with denominator less than k, so $a_{N+1}$ (and all following elements) must have a denominator larger than k. Since k was arbitrary, we have shown that the denominators of $\{a_n\}$ approach infinity. By definition of $f$, this means that $\{f(a_n)\}$, which is $\frac{1}{\textrm{denominator}}$ must go to zero.
\paragraph{}
Thus, for any $\{a_n\}$ composed of rational numbers that converges to $x$, $\{f(a_n)\} \to 0$. We showed earlier that we can merge any sequence of irrational numbers into the rational sequence (since the irrationals correspond to a 0 value of $f$) and still converge to zero. So for any sequence of real numbers $\{p_n\}$ that converges to some real $x$, $\{f(p_n)\}$ converges to zero.
\paragraph{}
We use this to show that $f(x)$ is continuous at every irrational point and has a simple discontinuity at every rational point. Suppose $x$ is irrational. Then $f(x) = 0$ by definition of the function, and $\lim_{t \to x} f(t) = 0 = f(x)$, so $f$ is continuous at $x$ by definition of continuity. Similarly, if $x$ is rational, then $f(x) \neq 0$, because $f(x)$ is the multiplicative inverse of the lowest terms denominator of x, which cannot be zero. But the limit at $x$ still exists and is zero: $\lim_{t \to x} f(t) = 0 \neq f(x)$. This is the definition of a discontinuity of the first kind.

\end{ppart}
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