First show that $\sup A + \sup B$ is an upper bound for $C$. This is true because every number in $C$ is of the form $a+b$ with $a\in A$ and $b\in B$. Since $\sup A$ is an upper bound for $A$ we have $a\leq \sup A$; similarly, we have $b\leq \sup B$. Therefore $a+b\leq \sup A + \sup B$, and we have shown that $\sup A+\sup B$ is an upper bound for $C$.
Next show that there is no smaller upper bound, i.e. suppose $m\lt \sup A+\sup B$ and show that $m$ is not an upper bound. To do this we define $r=(\sup A + \sup B)-m$. Since $m\lt \sup A + \sup B$, we know that $r\gt0$. The set $A$ must contain a number $a$ with $a\gt \sup A - \frac12 r$ because $\sup A$ is the least upper bound of $A$. Similarly, there is a $b\in B$ with $b\gt \sup B - \frac12 r$. Then $a+b\in C$ and $a+b\gt (\sup A + \sup B) - r = m$, which implies that $m$ is not an upper bound for $C$. The conclusion is that $\sup A+ \sup B$ is the lowest upper bound for $C$.
Let $m=\inf A$ where $A=\{x_1, x_2, x_3, \dots\}$. Since $0$ is a lower bound for $A$ the greatest lower bound $m$ for $A$ must exist. We will show that $m=0$. If $m$ were positive then, since $m$ is the greatest lower bound of $A$, the set $A$ must contain a number $x_n \in [m, \frac32m)$. But then $A$ also contains $x_{m+1}\leq \frac 23x_n \lt \frac23\cdot\frac32m = m$. Since $A$ apparently contains the number $x_{n+1}$ which is less than $m$, $m$ cannot be a lower bound for $A$, which contradicts the assumption $m=\inf A$. Hence we have shown that $\inf A=0$.
Since $0$ is the greatest lower bound for $A$, the number $1$ is not a lower bound, and therefore there is an $x_n$ with $x_n\lt 1$.
Assume $a_n\to 0$. To show $x_n\to p$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $|a_n|\lt \varepsilon$ for all $n\ge N$. This implies $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. Therefore $x_n\to p$.
Assume $x_n\to p$. To show $a_n\to 0$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. This implies $|a_n|\lt \varepsilon$ for all $n\ge N$. Therefore $a_n\to 0$.
We therefore have shown that any sequence of points in $E$ has a subsequence that converges to a point in $E$. This mean by definition that $E$ is compact.
a. $0\leq \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \lt \frac1{\sqrt n}$ so $x_n\to0$.
b. Cauchy sequences in $\R$ always converge, so $x_n=\sqrt n$ cannot be a Cauchy sequence. Alternatively, Cauchy sequences in any metric space are bounded, so $x_n=\sqrt n$ cannot be a Cauchy sequence.
In the following problems we let $X$ and $Y$ be a metric spaces, and write $d$ for the metric on both spaces.
$ d(f(p), f(q)) \leq C d(p, q)$ holds for all $p,q\in X$. $(*)$
Show that $f$ is continuous. (The condition $(*)$ is called a Lipschitz condition.)Since this holds for all $x,y$ we also have $f(y)-f(x) \leq d(y,x) = d(x,y)$, and thus
To prove that $F(x,y) = y-f(x)$ is continuous one can argue as follows: the two functions $(x,y)\to x $ and $ (x,y)\to y $ are the coordinate functions, so we know they are continuous (we proved this here).
The function $f(x)$, seen as a function on $\R^2$, is the composition of the coordinate function $ (x, y)\to x $ and the given function $x\to f(x)$, which is ggiven to be continuous. Therefore $ (x, y)\in\R^2 \to f(x) \in\R $ is continuous.
Finally, $y-f(x)$ is the difference of continuous functions, which is again continuous.
An approach that does work. Consider the function $h(x) = g(x)-f(x)$. We know that $h(a)=0$ and that $h'(x) = f'(x) - g'(x) \gt0$ for all $x\in(a, b)$. Apply the MVT to $h$ on the interval $[a,x]$: there is a $c\in(a, b)$ such that
Show that the series converges for all $x\in \R$. (Hint: to deal with the terms where $n\lt 0$ use that $|\sin\theta|\leq|\theta|$ holds for all $\theta\in\R$)
Also show that the graph of $f$ is self similar, in the sense that $f(3x) = 2f(x)$ holds for all $x\in\R$.
If $x\neq0$ then $1+nx^2\to\infty$ as $n\to\infty$, so $1/(1+nx^2)\to0$.
If $x=0$ then $f_n(x) = f_n(0) = 1$ for every $n\in\N$, so $f_n(0)\to 1$ as $n\to\infty$.
The limit function is therefore
One way to show this is to compute $\du(f_n, f) = \sup_{x\in\R}|f_n(x) - f(x)|$. It turns out that $\du(f_n, f)=1$ for all $n\in\N$, and therefore $\du(f_n, f)\not\to0$ as $n\to\infty$, which implies that $f_n$ does not converge uniformly.
In this problem there is an easier, indirect argument: all of the functions $f_n$ are continuous. If the convergence of $f_n$ to $f$ were uniform then the limit function $f$ would also have to be continuous. But $f$ is not continuous at $x=0$. Therefore $f_n$ does not converge uniformly to $f_n$ on $\R$.
In this case $f(x)=0$ for all $x\in[1,2]$, and the difference $|f_n(x)-f(x)|$ takes its maximum at $x=1$. Therefore $\du(f_n, f) = 1/(1+n)$, and $\du(f_n, f) \to 0$ as $n\to\infty$. This implies that $f_n(x)\to f(x)=0$ uniformly on $[1,2]$.
The same arguments, with the same choice of $M_n$ apply to the first question, i.e. the series converges absolutely for $|x|\leq a$ if $0\lt a\lt1$.
To prove uniform convergence on the interval $[-a, a]$, where $0\lt a\lt1$, we use the Weierstrass M–test. Choose $M_n = a^{n}$. Then we have for all $x$ with $|x|\leq a$ that $|x^{n^2}| \leq a^{n^2} \leq a^n$. We also have $\sum_{n=0}^\infty M_n =\sum a^n \lt\infty$. Weierstrass’ conditions are therefore met, and we can conclude that the series converges uniformly on $(-1, 1)$.
Each term is continuous (assuming that continuity of the Sine function as known), so therefore the sum of the series is also a continuous function.
To see that $A$ has no limit points, suppose $g\in\cC$ is a limit point of $A$, and consider the neighborhood $B_{1/2}(g)$. This neighborhood cannot contain more than one point from $A$, for if $f_n\in B_{1/2}(g)$ and $f_m\in B_{1/2}(g)$ then
If $B_{1/2}(g)$ contains one point $f_n\neq g$ from $A$, then let $r=\du(g, f_n)$; otherwise let $r=\frac12$. In either case $B_r(g)$ contains no points from $A$, except possibly $g$.