Analysis

[kanji]

1.Prove that for $x,y\in R^k$
(a.) $\left\Vert\,x\right\Vert\not= 0$ with equality only when $x=0$.
(b.) $\left\Vert\,\alpha x\right\Vert=\left|\,\alpha \right| \left\Vert\,x\right\Vert$ for all scalars $\alpha$.
(c.) $\left\Vert\,x+y\right\Vert\leq \left\Vert\,x\right\Vert+\left\Vert\,y\right\Vert$ and
$\left\Vert\,x-y\right\Vert\geq \left\Vert\,x\right\Vert-\left\Vert\,y\right\Vert$.
(d.) $\left\Vert\,x\right\Vert\leq \sum_{i = 1}^{k} \left\Vert\,x_i\right\Vert$.
(e.) $\left|\,x_i \right|\leq \left\Vert\,x\right\Vert\leq \sqrt{n}\left\Vert\,x_\infty \right\Vert$ for each $i=1,2,3,...,k$.

[M@gpie]

ไม่แน่ใจว่าเข้าใจโจทย์ผิดรึเปล่า แต่ให้แสดงว่า $\|x\| = 0 \; \Leftrightarrow \; x=0, \; \; \forall x \in \mathbb{R}^n$ ใช่รึเปล่าครับผม? ขอทำแบบนี้ละกัน
(a) ($\Rightarrow$) Let $x\in \mathbb{R}^n$ with Euclidean norm. It is easy to show that \[ \|x\|^2 = x_1^2+x_2^2+...+x_n^2 = 0 \Rightarrow x_i = 0, \; \; \forall i\in \{1,2, ..., n\} \]
($\Leftarrow$) is obvious.
(b) \[ \| \alpha \cdot x\| = \sqrt{(\alpha x_1)^2 + (\alpha x_2)^2+ ... +(\alpha x_n)^2} = |\alpha |\|x\| \]
(c) To show triangle inequality for $x\in \mathbb{R}^n$, we have to use Cauchy-Swarchz inequality such as $|\langle x,y\rangle | \leq \|x\| \|y\| $
Hence, \[ \| x+y \|^2 = \|x\|^2+ 2\langle x,y\rangle + \|y\|^2 \leq (\|x\|+\|y\|)^2.\]
The other inequality follows from the above result. Therefore, \[ \|x\| = \|x-y+y\| \leq \|x-y\| +\|y\|\]

(e) It is clear that \[ x_i^2 \leq x_1^2+x_2^2+...+x_n^2 \leq n (\sup_{1\leq i \leq n}|x_i|)^2\]
Hence, \[ |x_i| \leq \|x\| \leq \sqrt{n}\|x\|_{\infty}\]

[kanji]

2.Let $x_n\in R$ and suppose that there is an $M\in R$ such that $\left|\,x\right|\leq M$ for $n\in N$. Prove that $s_n=sup\left\{\,x_n,x_{n+1},...\right\} $ defines a real number for each $n\in N$ and $s_1\geq s_2\geq ...$.
3.Give an example of a set of rational numbers which is bounded but does not have a rational supremum.
4.Give an example of a set of rational numbers that has a rational supremum.

[M@gpie]

2. For each $n \in \mathbb{N}$, we have $s_{n+1}= \{ x_k : k\geq n+1\} \subset s_n=\{ x_k : k\geq n \}.$
Notice that if $A,B$ are above bounded sets such that $A\subseteq B$, then $\sup A \leq \sup B.$ This implies that ${\displaystyle s_n = \sup_{k\geq n}x_k }$ is a decreasing sequence as required.

3. By density theorem, we can find a sequence which converges to $\sqrt{2}$
Then we define a sequence of rational number such that \[ 1, \; \; 1.4, \; \; 1.41, \; \; 1.414, \; \; 1.414,\; \; 1.4142, \; \; 1.41421, ... \]
We can see that supermum of this sequence is $\sqrt{2}$ which is not rational.

4. Choose $a_n = \frac{1}{n}$, then ${\displaystyle \sup_{n\in \mathbb{N}} a_n = 1 }$

[kanji]

ข้อ e น่าจะเป็น $(suplx_il)^2$ ใช่ไหมครับ

[M@gpie]

แก้ให้แล้วครับผม

[kanji]

อีกชุด ครับ
1.Prove that $A$ is dense in $X$ if and only if $\overline{A} =X$. where $\overline{A} $ = closure of $A$.
2.Prove that $Q$ and $R-Q$ are dense in $R$.
3.A point $p$ is an element of $\overline{A}$ if and only if for every positive real number $r$, $N_r(p)\cap A\not= \emptyset $.
4.Prove that $X$ is connrcted if and only if $X$ has exactly two subsets which are both open and closed in $X$.
5.Let $X$ be any nonempty set and $d$ the discrete metric on $X$. Prove that every subset of $X$ is open and every subset is closed.

[passer-by]

2. HINT:

let y be arbitrary in R. Consider open neighborhood $(y- \epsilon, y+\epsilon)$ and using the fact that there's at least 1 rational number between any two real numbers. This proof follows from topological version of dense set.

About the fact I claim above, It's good for future pure mathematician that you should prove before using.

For irrational case, it follows similarly and adjust something.

But if you hate topological version, you might use sequential version of dense set. For any arbitrary $a \in R$ , consider $ (a- \frac{1}{n}, a+\frac{1}{n}) $ for each n and applying the same fact.


5. HINT:
Let $ a \in R_d $ , consider open ball radius 1 ,say, B[a;1] = {a} to conclude that every subset is open. For proving closed, it's automatically from theorem that complement of open set is closed set.

[M@gpie]

ตรวจให้ด้วยนะคร้าบ พี่ๆน้องๆ ทั้งหลาย
1. $(\Rightarrow )$ Suppose $A$ is dense in $X$,
then for any $\epsilon >0$ and any $x\in X$, there exists $a\in A$ such that $|x-a| < \epsilon$.
Now, let $x\in X$. For any $n\in \mathbb{N}$, there is $a_n \in A$ such that $|a_n-x|<\frac{1}{n}$. Hence, $x\in \overline{A}$.
This implies that $\overline{A}=X$
$(\Leftarrow )$ Suppose $\overline{A}=X$.
Notice that if $x\in \overline{A}$, then there is a sequence $(a_n)$ which converges to $x$.
Thus, for any $\epsilon > 0$ and $x\in X$, there is $a_{N} \in A$ such that $|a_{N} -x| < \epsilon$.
This shows that $A$ is dense in $X$.

2. It follows from the density theorem of the real number.

3. เป็น Definition รึเปล่าครับ??

[nooonuii]

4. $(\Rightarrow)$ Suppose there is a clopen set $A\neq \emptyset,X$. What can you say about $X-A$ ?

$(\Leftarrow)$ Let $f:X\to D$ be a discrete valued map ($D$ has discrete topology). Pick any $c\in f(X)$. Then show that $f^{-1}(c)$ is nonempty closed and open in $X$. Conclude that $f$ is constant.

Then use the fact that

$X$ is connected iff any discrete valued map $f:X\to D$ is constant.

For this part you can use the same argument as in the first part too. I just give you another way to prove that $X$ is connected. This approach is much easier in many situations.

[kanji]

ข้อ 3 ทำยังไงดีครับ

[nooonuii]

ข้อสามควรบอกนิยามของ closure มาก่อนครับ เพราะมีการใช้สิ่งที่เราต้องการพิสูจน์เป็นนิยามในหนังสือบางเล่มครับ

[kanji]

Definition :
If $X$ is a metric space, if $A\subseteq X$, and if $A'$ denotes the set of all limit points of $A$ in $X$, then the closure of $A$ is the set $\bar A= A \cup A'$.

[nooonuii]

3. $(\Rightarrow)$ Consider two cases.

I. $p\in A$. Obvious.
II. $p\in A'$. Follow directly from definition of a limit point.

$(\Leftarrow)$. Suppose $p\not\in A$. Then use this assumption to show that $p$ is a limit point of $A$.

[mercedesbenz]

ข้อความเดิมเขียนโดยคุณ passer-by

2. HINT:



5. HINT:
Let $ a \in R_d $ , consider open ball radius 1 ,say, B[a;1] = {a} to conclude that every subset is open. For proving closed, it's automatically from theorem that complement of open set is closed set.

ช่วยขยาย Hint ให้อีกสักหน่อยได้ไหมครับ พอดีทำแล้วยังคิดไม่ออก