Differences

This shows you the differences between two versions of the page.

--- math104-f21:hw7 [2021/10/08 17:25]
pzhou
+++ math104-f21:hw7 [2026/02/21 14:41] (current)
@@ Line 15: / Line 15: @@
 Optional. Let $(X, d)$ be a metric space, $k \geq 1$ be an integer. Let $Conf_k(X) = \{ S \In X, |S|=k\}$, i.e., an element in $Conf_k(X)$ is subset  $S \In X$ consisting of $k$ points. For example, let $X = \R$, $k=2$, then $S = \{ 2, -1.1 \}$ is an element in $Conf_2(X)$. Can you put a metric on $Conf_k(X)$ using $d$?
+====== Solution ======
+. Consider the metric space $(X = \Z, d_X(x,y) = |x-y|)$. Write down the open ball $B_3 (2)$, $B_{1/2}(2)$. Is the subset $\{2\}$ open in $X$? Is it closed in $X$? Explain.
+  * $B_3(2) = \{0, 1, 2, 3, 4\}$
+  * $B_{1/2}(2) = \{2\}$
+  * $\{2\}$ is open, since it contains open ball $B_{1/2}(2)$. It is closed, since any singleton is closed.
+. Consider the metric space $(X = [0,1) \In \R, d_X(x,y) = |x-y|)$. Write down the open ball $B_{3} (0)$, $B_{1/3}(0)$. Is the subset $[0,1)$ open in $X$? closed in $X$? Is the subset $[1/3, 1)$ open in $X$? closed in $X$?
+  * $B_3(0) = [0,1) $
+  * $B_{1/3}(0) = [0, 1/3)$
+  * Since $[0,1)$ is $X$, hence $X$ is open in $X$, is closed in $X$.
+  * $[1/3, 1)$ is not open in $X$, since the point $1/3 \in [1/3, 1)$ does not have any open neighborhood in $[1/3,1)$. It is closed in $X$, since the complement $[0,1/3)$ is open.
+. Construct a subset $E \In [0,1]$, such that the limit points of $E$ is $E' = \{0, 1\}$. Optional:  Is it possible to construct $E$, such that   $E' = \{0, 1, 1/2, 1/3, \cdots \}$?
+  * $E = \{ 1/n | n =1,2, \cdots \} \cup \{ 1- 1/n | n =1,2, \cdots \} $, then $E' = \{0,1\}$.
+  * Optional problem: yes it is possible. For any integer $n \geq 2$, let $r_n = 1/(n-1) - 1/n$. Let $E_n = \{ 1/n + r_n 1/m \mid m = 1,2,\cdots \}$. Then $E_n \In (1/n, 1/(n-1))$, and $E_n' = \{1/n\}$. Let
+$$ E = \cup_{n=2}^\infty E_n \bigcup  \{ 1- 1/m | m =1,2, \cdots \} $$
+The last factor is added to get limit point $1$.
+. Let $X =\R$, let $d(x,y) = \sqrt{|x-y|}$. Is $d(x,y)$ a distance function on $X$?
+Yes, just need to check triangle inequality.
+$$ d(x,y) + d(y,z) \geq d(x,z) $$
+$$\Leftrightarrow (d(x,y) + d(y,z))^2 \geq d(x,z)^2 $$
+$$ \Leftrightarrow |x-y| + |y-z| + 2 \sqrt{ |x-y| |y-z|} \geq |x-z| $$
+$$ \Leftarrow |x-y| + |y-z| \geq |x-z| $$
+where the last step is due to $2 \sqrt{ |x-y| |y-z|}  \geq 0$.
+. Let $X$ be a metric space, and $E \In X$ be any subset. Prove that $\overline{E^c} = (E^o)^c$, where $E^o$ means the set of interior points in $E$.
+We know
+$$ \overline{E^c} = \bigcap \{ K \mid  K \In X \text{ is closed}, E^c \In K \} = \bigcap \{ K \mid K \In X \text{ is closed}, K^c \In E \} = \bigcap \{ K \mid K^c \In X \text{ is open}, K^c \In E \} $$
+and
+$$ E^o = \bigcup \{ F \mid  F \In X \text{ is open}, F \In E \} $$
+Hence
+$$ (E^o)^c = \bigcap \{ F^c \mid  F \In X \text{ is open}, F \In E \} = \bigcap \{ K \mid K^c \In X \text{ is open}, K^c \In E \} = \overline{E^c} $$
+Optional. Let $(X, d)$ be a metric space, $k \geq 1$ be an integer. Let $Conf_k(X) = \{ S \In X, |S|=k\}$, i.e., an element in $Conf_k(X)$ is subset  $S \In X$ consisting of $k$ points. For example, let $X = \R$, $k=2$, then $S = \{ 2, -1.1 \}$ is an element in $Conf_2(X)$. Can you put a metric on $Conf_k(X)$ using $d$?
+See [[https://en.wikipedia.org/wiki/Hausdorff_distance | Hausdorff distance ]]

Lecture Notes

User Tools

Site Tools

Differences

Page Tools