实分析代写| Topology of the Real Number Line代写|实轴的拓扑代写|Math115A代写

Proof . (a) $[a, b]=\bigcap_{n=1}^{\infty}(a-1 / n, b+1 / n)$.
(b) Let $N$ be large enough so that $b-a>\frac{2}{N}$ (so that we do not get empty set in the union, but it is also fine if you include many emtpy sets). Then $(a, b)=\bigcup_{n=N}^{\infty}[a+1 / n, b-1 / n]$.

Proof . (a) Let $x \in U,$ then by definition there is some $\lambda$ such that $x \in U_{\lambda} .$ Since $U_{\lambda}$ is open we have an open interval $I_{x}$ such that $x \in I_{x} \subset U_{\lambda} \subset U,$ so $U$ is open.
(b) Let $x \in U,$ then $x \in U_{i}$ for each $i=1, \ldots, n .$ By definition there are open intervals $I_{1}, \ldots I_{n}$ such that $x \in I_{i} \subset U_{i} .$ Without loss of generality, and after possibly shrinking $I_{i},$ we can assume $I_{i}=\left(x-a_{i}, x+a_{i}\right)$ for $a_{i} \in \mathbb{R}$. Now choose a real number $a<\min \left\{a_{1}, \ldots, a_{n}\right\},$ it follows that $(x-a, x+a) \subset I_{i} \subset U_{i}$ for each $i=1, \ldots, n,$ and hence $(x-a, x+a) \subset \bigcap_{i=1}^{n} U_{i}=U,$ so $U$ is
open.
(c) The example is the same as Problem 1 (a). We will show that the closed interval (as the name suggests) $[a, b]$ is not open. In fact, any open interval around $b$ will not be fully contained in $[a, b]$. Indeed, any interval around $b$ contains an interval of the form $(b-\varepsilon, b+\varepsilon)$ for some $\varepsilon>0,$ and it is clear that $b+\varepsilon / 2 \notin[a, b]$.

Proof . Since $\mathbb{Q}$ is dense in $\mathbb{R}$, given any $x \in U$ and an interval $(x-a, x+a) \subset U, \mathbb{Q} \cap(x-a, x+a)$ is not empty, so there is at least some rational number in it.

Proof . (a) We claim that the limit is $1 .$ Indeed, given $\varepsilon>0,$ let $N>0$ be such that $\frac{1}{N}<\varepsilon,$ then for all $n>N$ we have
$$
\left|\frac{n}{n+1}-1\right|=\frac{1}{n+1}<\frac{1}{N}<\varepsilon
$$
So the limit is 1 by definition.
(b) We claim that the limit does not exist. Suppose for a contradiction that $L \in \mathbb{R}$ is the limit of the sequence. We must exhibit an $\varepsilon$ such that there is no $N \in \mathbb{N}$ such that
$$
\left|\frac{(-2)^{n}}{n^{2}}-L\right|<\varepsilon
$$
for all $n>N$. We will show this using $\varepsilon=1$. Let $n$ be even and by triangle in equality we have
$$
\left|\frac{(-2)^{n}}{n^{2}}-L\right| \geq \frac{2^{n}}{n^{2}}-L
$$
Now since $2^{n}$ grows much faster than $n^{2}$, given any $N \in \mathbb{N},$ we can choose an even number $n>N$ large so that
$$
\frac{2^{n}}{n^{2}}>L+2
$$
and it follows
$$
\left|\frac{(-2)^{n}}{n^{2}}-L\right|>L+2-L>2>1
$$
a contradiction, so the limit must not be equal to $L$ for any $L \in \mathbb{R} .$ Hence the limit must not exist.

Proof .

Without loss of generality we may suppose $\left\{x_{n}\right\}$ is nondecreasing (otherwise consider $\left\{-x_{n}\right\}$ ). Suppose for a contradiction that there is some $N>k$ such that $x_{N}>x_{k}$ (since the sequence is nondecreasing $\left.x_{N} \geq x_{k}\right) .$ Let $\varepsilon=x_{N}-x_{k},$ it follows from monotonicity that
$$
\left|x_{n}-x_{k}\right|=x_{n}-x_{k} \geq x_{N}-x_{k}>\varepsilon
$$
for every $n \geq N,$ so $x_{k}$ clearly cannot be the limit of the sequence, a contradiction.

Proof . (a) Let $I_{n}=\left[a_{n}, b_{n}\right] .$ The nested condition ensures that $\left\{a_{n}\right\}$ is a monotone nondecreasing sequence. Note also that $\left\{a_{n}\right\}$ is bounded above by $b_{1},$ so by the monotone convergence theorem we have $a_{n} \rightarrow a$ for some $a \in \mathbb{R} .$ We claim that $a \in \bigcap_{n=1}^{\infty} I_{n} .$ Note that it suffices to show that $a \leq b_{n}$ for every $n$. Suppose for a contradiction that $a>b_{n 0}$ for some $n_{0} .$ Let $\varepsilon=a-b_{n 0} .$ Since $a_{n} \rightarrow a,$ there is $N \in \mathbb{N}$ such that $n>N$ implies $a-a_{n}<\varepsilon=a-b_{n 0} .$ On the other hand, since the intervals are nested we have $a-b_{n 0}<a-b_{n}$ for every $n>n_{0} .$ Thus choosing $n>\max \left\{N, n_{0}\right\}$ we have
$$
a-a_{n}<\varepsilon=a-b_{n_{0}}<a-b_{n} \Longrightarrow b_{n}<a_{n}
$$
a contradiction, since the intervals are assumed to be nonempty. This proves $a \leq b_{n}$ for every $n$ and so $a \in I_{n}$ for every $n,$ which in turn implies $a \in \bigcap_{n=1}^{\infty} I_{n}$.
(b) Consider $J_{n}=(0,1 / n),$ then $\bigcap_{n=1}^{\infty} J_{n}=\emptyset$.