### 数学代写|实变函数作业代写Real analysis代考|Bolzano – Weierstrass Results

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## 数学代写|实变函数作业代写Real analysis代考|Bounded Sequences with a Finite Range

We have already looked at sequences with finite ranges. Since their range is finite, they are bounded sequences. We also know they have subsequences that converge which we have explicitly calculated. If the range of the sequence is a single value, then we know the sequence will converge to that value and we now know how to prove convergence of a sequence. Let’s formalize this into a theorem. But this time, we will argue more abstractly. Note how the argument is still essentially the same.
Theorem 4.1.1 A Sequence with a Finite Range Diverges Unless the Range is One Value
Let the sequence $\left(a_{n}\right)$ have a finite range $\left{y_{1}, \ldots, y_{p}\right}$ for some positive integer $p \geq 1$. If $p=1$, the sequence converges to $y_{1}$ and if $p>1$, the sequence does not converge but there is a subsequence $\left(a_{n_{k}^{i}}\right)$ which converges to $y_{i}$ for each $y_{i}$ in the range of the sequence.
Proof 4.1.1
If the range of the sequence consists of just one point, then $a_{n}=y_{1}$ for all $n$ and it is easy to see $a_{n} \rightarrow y_{1}$ as given $\epsilon>0,\left|a_{n}-y_{1}\right|=\left|y_{1}-y_{1}\right|=0<\epsilon$ for all $n$ which shows convergence.

If the range has $p>1$, let a be any number not in the range and calculate $d_{i}=\left|a-y_{i}\right|$, the distance from a to each point $y_{i}$ in the range. Let $d=(1 / 2) \min \left{d_{1}, \ldots, d_{p}\right}$ and choose $\epsilon=d$. Then $\left|a_{n}-a\right|$ takes on $p$ values, $\left|y_{i}-a\right|=d_{i}$ for all $n$. But $d_{i}>d$ for all $i$ which shows us that $\left|a_{n}-a\right|>\epsilon$ for all $n$. A little thought then shows us this is precisely the definition of the sequence $\left(a_{n}\right)$ not converging to $a$.

If $a$ is one of the range values, say $a=y_{i}$, then the distances we defined above are positive except $d_{i}$ which is zero. So $\left|a_{n}-y_{i}\right|$ is zero for all indices $n$ which give range value $y_{i}$ but positive for all other range values. Let $\epsilon=d=(1 / 2) \min {j \neq i}\left|d{i}-d_{j}\right|$. Then, for any index $n$ with $a_{n} \neq y_{i}$, we have $\left|a_{n}-y_{i}\right|=\left|y_{j}-y_{i}\right|=d_{j}>d$ for some $j$. Thus, no matter what $N$ we pick, we can always find $n>N$ giving $\left|a_{n}-y_{i}\right|>\epsilon$. Hence, the limit can not be $y_{i}$. Since this argument works for any range value $y_{i}$, we see the limit value can not be any range value.

To make this concrete, say there were 3 values in the range, $\left{y_{1}, y_{2}, y_{3}\right}$. If the limit was $y_{2}$, let $\epsilon=d=(1 / 2) \min \left{\left|y_{1}-y_{2}\right|,\left|y_{3}-y_{2}\right|\right}$. Then,

$$\left|a_{n}-y_{2}\right|= \begin{cases}\left|y_{2}-y_{1}\right|>d=\epsilon, & a_{n}=y_{1} \ \left|y_{2}-y_{2}\right|=0, & a_{n}=y_{2} \ \left|y_{3}-y_{2}\right|>d,=\epsilon & a_{n}=y_{3}\end{cases}$$
Given any $N$ we can choose $n>N$ so that $\left|a_{n}-y_{2}\right|>\epsilon$. Hence, the limit can not be $y_{2}$. Note how this argument is much more abstract than our previous ones.

Comment 4.1.1 For comvenience of exposition (cool phrase…) let’s look at the range value $y_{1}$. The sequence has a block which repeats and inside that block are the different values of the range $y_{i}$. There is a first time $y_{1}$ is present in the first block. Call this index $n_{1}$. Let the block size by $Q$. Then the next time $y_{1}$ occurs in this position in the block is at index $n_{1}+Q$. In fact, $y_{1}$ occurs in the sequence at indices $n_{1}+j Q$ where $j \geq 1$. This defines the subsequence $a_{n_{1}}+j Q$ which converges to $y_{1}$. The same sort of argument can be used for each of the remaining $y_{i}$.

## 数学代写|实变函数作业代写Real analysis代考|Sequences with an Infinite Range

We are now ready for our most abstract result so far.
Theorem 4.2.1 Bolzano – Weierstrass Theorem
Every bounded sequence has at least one convergent subsequence.
Proof 4.2.1
As discussed, we have already shown a sequence with a bounded finite range always has convergent subsequences. Now we prove the case where the range of the sequence of values $\left{a_{1}, a_{2} \ldots,\right$,$} has$ infinitely many distinct values. We assume the sequences start at $n=k$ and by assumption, there is a positive number $B$ so that $-B \leq a_{n} \leq B$ for all $n \geq k$. Define the interval $J_{0}=\left[\alpha_{0}, \beta_{0}\right]$ where $\alpha_{0}=-B$ and $\beta_{0}=B$. Thus at this starting step, $J_{0}=[-B, B]$. Note the length of $J_{0}$, denoted by $\ell_{0}$ is $2 B$.

Let $\mathcal{S}$ be the range of the sequence which has infinitely many points and for convenience, we will let the phrase infinitely many points be abbreviated to IMPs.
Step 1:
Bisect $\left[\alpha_{0}, \beta_{0}\right]$ into two pieces $u_{0}$ and $u_{1}$. That is the interval $J_{0}$ is the union of the two sets $u_{0}$ and $u_{1}$ and $J_{0}=u_{0} \cup u_{1}$. Now at least one of the intervals $u_{0}$ and $u_{1}$ contains IMPs of $\mathcal{S}$ as otherwise each piece has only finitely many points and that contradicts our assumption that $\mathcal{S}$ has IMPS. Now both may contain IMPS so select one such interval containing IMPS and call it $J_{1}$. Label the endpoints of $J_{1}$ as $\alpha_{1}$ and $\beta_{1}$; hence, $J_{1}=\left[\alpha_{1}, \beta_{1}\right] .$ Note $\ell_{1}=\beta_{1}-\alpha_{1}=\frac{1}{2} \ell_{0}=B$ We see $J_{1} \subseteq J_{0}$ and
$$-B=\alpha_{0} \leq \alpha_{1} \leq \beta_{1} \leq \beta_{0}=B$$
Since $J_{1}$ contains IMPS, we can select a sequence value $a_{n_{1}}$ from $J_{1}$.
Step 2:
Now bisect $J_{1}$ into subintervals $u_{0}$ and $u_{1}$ just as before so that $J_{1}=u_{0} \cup u_{1}$. At least one of $u_{0}$ and $u_{1}$ contain IMPS of $\mathcal{S}$.

Choose one such interval and call it $J_{2}$. Label the endpoints of $J_{2}$ as $\alpha_{2}$ and $\beta_{2}$; hence, $J_{2}=$ $\left[\alpha_{2}, \beta_{2}\right]$. Note $\ell_{2}=\beta_{2}-\alpha_{2}=\frac{1}{2} \ell_{1}$ or $\ell_{2}=(1 / 4) \ell_{0}=\left(1 / 2^{2}\right) \ell_{0}=(1 / 2) B$. We see $J_{2} \subseteq J_{1} \subseteq J_{0}$ and
$$-B=\alpha_{0} \leq \alpha_{1} \leq \alpha_{2} \leq \beta_{2} \leq \beta_{1} \leq \beta_{0}=B$$

## 数学代写|实变函数作业代写Real analysis代考|Extensions to $\Re^{2}$

We can extend our arguments to bounded sequences in $\mathfrak{R}^{2}$. We haven’t talked about it yet, but given a sequence in $\Re^{2}$, the elements of the sequence will be vectors. The sequence then will be made up of vectors.
$$\left(\left(x_{n}\right)\right)=\left[\begin{array}{l} x_{1, n} \ x_{2, n} \end{array}\right]$$
We would say the sequence converges to a vector $\boldsymbol{x}$ is for all $\epsilon>0$, there is an $N$ so that
$$n>N \Longrightarrow\left|\boldsymbol{x}{n}-\boldsymbol{x}\right|{2}<\epsilon$$
where we measure the distance between two vectors in $\Re^{2}$ using the standard Euclidean norm, here called $|\cdot|_{2}$ defined by
$$\left|\boldsymbol{x}{n}-\boldsymbol{x}\right|{2}=\sqrt{\left(x_{1, n}-x_{1}\right)^{2}+\left(x_{2, n}-x_{2}\right)^{2}}$$
where $\boldsymbol{x}=\left[x_{1}, x_{2}\right]^{\prime}$. This sequence is bounded if there is a positive number $B$ so that $\left|\boldsymbol{x}{\boldsymbol{n}}\right|{2}<B$ for all appropriate $n$. We can sketch the proof for the case where there are infinitely many vectors in this sequence which is bounded.
Theorem 4.2.2 The Bolzano – Weierstrass Theorem in $\Re^{2}$
Every bounded sequence of vectors in $\mathrm{R}^{2}$ with an infinite range has at least one convergent subsequence.
Proof 4.2.3
We will just sketch the argument. Since this sequence is bounded, there are positive numbers $B_{1}$ and $B_{2}$ so that
$$-B_{1} \leq x_{1 n} \leq B_{1} \quad \text { and } \quad-B_{2} \leq x_{2 n} \leq B_{2}$$
The same argument we just used for the Bolzano – Weierstrass Theorem in $\Re$ works. We bisect both edges of the box to create 4 rectangles. At least one must contain IMPs and we choose one that does. Then at each step, continue this subdivision process always choosing a rectangle with IMPs. Here are a few of the details. We start by labeling the initial box by

$$J_{0}=\left[-B_{1}, B_{1}\right] \times\left[-B_{2}, B_{2}\right]=\left[\alpha_{0}, \beta_{0}\right] \times\left[\delta_{0}, \gamma 0\right]$$
We pick a first vector $x_{n_{0}}$ from the initial box. The area of this rectangle is $A_{0}=\left(\beta_{0}-\alpha_{0}\right)\left(\gamma_{0}-\delta_{0}\right)$ and at the first step, the bisection of each edge leads to four rectangles of area $A_{0} / 4$. At least one of these rectangles contains IMPs and after our choice, we label the new rectangle. $J_{1}=\left[\alpha_{1}, \beta_{1}\right] \times$ $\left[\delta_{1}, \gamma_{1}\right]$ and pick a vector $\boldsymbol{x}{n{1}}$ different from the first one from $J_{0}$.
At this point, we have
\begin{aligned} &-B_{1}=\alpha_{0} \leq \alpha_{1} \leq \beta_{1} \leq \beta_{0}=B_{1} \ &-B_{2}=\delta_{0} \leq \delta_{1} \leq \gamma_{1} \leq \gamma_{0}=B_{2} \end{aligned}
and after $p$ steps, we have
\begin{aligned} &-B_{1}=\alpha_{0} \leq \alpha_{1} \leq \ldots \alpha_{p} \leq \beta_{p} \leq \ldots \leq \beta_{1} \leq \beta_{0}=B_{1} \ &-B_{2}=\delta_{0} \leq \delta_{1} \leq \ldots \delta_{p} \leq \gamma_{p} \leq \ldots \leq \gamma_{1} \leq \gamma_{0}=B_{2} \end{aligned}
with the edge lengths $\beta_{p}-\alpha_{p}$ and $\gamma_{p}-\delta_{p}$ going to zero as $p$ increases, As before, we pick a value $\boldsymbol{x}{n{p}}$ different from the ones at previous stages. Similar arguments show that $\alpha_{p} \rightarrow \alpha$ and $\beta_{p} \rightarrow \beta$ with $\alpha=\beta$ and $\delta_{p} \rightarrow \delta$ and $\gamma_{p} \rightarrow \gamma$ with $\delta=\gamma$. We find also $\boldsymbol{x}{n{p}} \rightarrow[\alpha, \delta]^{\prime}$ which gives us the result. The convergence arguments here are indeed a bit different as we have to measure distance between two vectors using $\left|_{1}\right|_{2}$ but it is not too difficult to figure it out.

## 数学代写|实变函数作业代写Real analysis代考|Bounded Sequences with a Finite Range

|一种n−是2|={|是2−是1|>d=ε,一种n=是1 |是2−是2|=0,一种n=是2 |是3−是2|>d,=ε一种n=是3

## 数学代写|实变函数作业代写Real analysis代考|Sequences with an Infinite Range

−乙=一种0≤一种1≤b1≤b0=乙

−乙=一种0≤一种1≤一种2≤b2≤b1≤b0=乙

## 数学代写|实变函数作业代写Real analysis代考|Extensions to ℜ2

((Xn))=[X1,n X2,n]

n>ñ⟹|Xn−X|2<ε

|Xn−X|2=(X1,n−X1)2+(X2,n−X2)2

−乙1≤X1n≤乙1 和 −乙2≤X2n≤乙2

Ĵ0=[−乙1,乙1]×[−乙2,乙2]=[一种0,b0]×[d0,C0]

−乙1=一种0≤一种1≤b1≤b0=乙1 −乙2=d0≤d1≤C1≤C0=乙2

−乙1=一种0≤一种1≤…一种p≤bp≤…≤b1≤b0=乙1 −乙2=d0≤d1≤…dp≤Cp≤…≤C1≤C0=乙2

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