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--- math104-f21:hw10 [2021/10/30 09:28]
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 . Equivalence of metrics on $\R^2$. Recall that we can equip $\R^2$ with the usual Euclidean metric $d_2(\vec x, \vec y) = \sqrt{|x_1 - y_1|^2 + |x_2- y_2|^2}$, or with the metric
-$d_\infty(\vec x, \vec y) = \max(|x_1 - y_1|, |x_2 - y_2|)$. Prove that $d_1$ and $d_\infty$ are equivalent.
+$d_\infty(\vec x, \vec y) = \max(|x_1 - y_1|, |x_2 - y_2|)$. Prove that $d_2$ and $d_\infty$ are equivalent.
 . Equivalent metrics induces the same topology. Let the set $X$ be equipped with two **equivalent** metrics $d_1, d_2$.  Consider two types of balls
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 $$ c_1 d_2(0, \vec a) < d_\infty(0, \vec a) < c_2 d_2 (0, \vec a). $$
+====== Solution ======
+. Equivalence of metrics on $\R^2$. Recall that we can equip $\R^2$ with the usual Euclidean metric $d_2(\vec x, \vec y) = \sqrt{|x_1 - y_1|^2 + |x_2- y_2|^2}$, or with the metric
+$d_\infty(\vec x, \vec y) = \max(|x_1 - y_1|, |x_2 - y_2|)$. Prove that $d_2$ and $d_\infty$ are equivalent.
+For any real numbers $a,b \in \R$, let $A = \max\{|a|,|b|\}$. We have
+$$ \sqrt{a^2 + b^2} \leq \sqrt{A^2 + A^2} = \sqrt{2} A $$
+and
+$$ \sqrt{a^2+b^2} \geq \sqrt{A^2} = A. $$
+. Equivalent metrics induces the same topology. Let the set $X$ be equipped with two **equivalent** metrics $d_1, d_2$.  Consider two types of balls
+$$ B_r^{d_i}(p) = \{ p' \in X \mid d_i(p,p') < r\} $$
+  * prove that for any $p \in X$, $r > 0$, and any $q \in B_r^{d_1}(p)$, there exists $\epsilon>0$, such that $B_\epsilon^{d_2}(q) \subset B_r^{d_1}(p)$.
+  * prove that if a subset $U \subset X$ is open with the $d_1$-metric, then it is open in the $d_2$-metric.
+Since $d_1, d_2$ are equivalent, we then have $c_1, c_2>0$, such that $c_1 d_1 \leq d_2 \leq c_2 d_1$.
+(a) Let $a = r - d_1(p,q)$, then the ball $B^{d_1}_a(q) \subset B^{d_1}_r(p)$. Since for any $x \in X$, we have $d_1(x,q) \leq d_2(x,q) / c_1$. Hence, if $d_2(x,q) < c_1 a$, then $d_1(x,q) < a$. Thus, if $\epsilon = c_1 a$, we have
+$$ B^{d_2}_{\epsilon} (q) \subset B^{d_1}_a(q) \subset B^{d_1}_r(p). $$
+(b) If $U$ is open with $d_1$ metric, then for every point $x \in U$, there exists a ball $B^{d_1}_r(x) \subset U$. By part (a), we there also exists $B^{d_2}_\epsilon(x) \subset B^{d_1}_r(x) \subset U$. Hence $U$ is open in the $d_2$-metric sense.
+. Addition on $\R^2$ is continuous. Let $f:\R^2 \to \R$ be given by $f(x,y) = x+y$. Prove that $f$ is continuous. You can use either one of the following definition for continuous functions
+  * (metric space sense) $f: (X, d_X) \to (Y, d_Y)$ is continuous if for any $x \in X$, and any $\epsilon>0$, there exists $\delta>0$, such that $f (B_\delta(x)) \subset B_\epsilon(f(x))$.
+  * (topological sense) $f: X \to Y$ is open, if for any open subset $V$ in $Y$, $f^{-1}(V)$ is open.
+Proof in the metric sense: We are going to use $d_\infty$-metric on $\R^2$. For any $(x_0, y_0) \in \R^2$,  for any $\epsilon>0$, we need to show that there exists $\delta>0$, such that for any $|x - x_0| < \delta, |y-y_0| < \delta$, we have $|f(x,y) - f(x_0, y_0)| < \epsilon$. We can let $\delta = \epsilon/2$, then
+$$ |f(x,y) - f(x_0, y_0)| = |x + y - (x_0 + y_0) | \leq |x - x_0| + |y - y_0| \leq \delta + \delta = \epsilon. $$
+. Let $X$ be a metric space $A \subset X$ be any subset. Define the distance function to $A$ as
+$$ d_A(p) = \inf \{ d(p,q) \mid q \in A \} $$
+Prove that $d_A$ is a continuous function on $X$, and show that $d_A(p)=0$ if and only if $p \in \bar A$.
+To show that $d_A$ is continuous, we need to show that for any $p \in X$ and $\epsilon >0$, we have $\delta > 0$, such that for any $q \in X$ with $d(p,q) < \delta$, we have $|d_A(p) - d_A(q)|<\epsilon$.
+Note that
+$$ d_A(q) = \inf \{ d(x,q) \mid x \in A \} \leq \inf \{ d(x,p) + d(p,q) \mid x \in A \} = (\inf \{ d(x,p) \mid x \in A \})  + d(p,q) = d_A(p) + d(p,q) $$
+similarly, we have
+$$ d_A(p) \leq d_A(q) + d(p,q).$$
+Hence
+$$ | d_A(p) - d_A(q) | \leq d(p,q) $$
+Thus, taking $\delta = \epsilon$ would work.
+For the second part, first note that $d_A^{-1}(0)$ is a closed subset in $X$ (since $\{0\}$ is closed, and $d_A$ is continuous, hence the preimage is closed) and contains $A$, hence $\bar A \subset d_A^{-1}(0)$. Thus if $p \in \bar A$, then $d_A(p)=0$.  And if $p \notin \bar A$, then there exists $B_r(p) \cap \bar A = \emptyset$, this implies for any $x \in A$, $x \notin B_r(p)$, thus $d(x,p) \geq r$, hence $d_A(p) \geq r$, in particular $d_A(p) \neq 0$.
+. Let $P = \{\vec a = ( a_1, a_2, \cdots, ) \mid a_i \in \R, a_N=0 \text{ for } N \gg 0 \}$ be the set of real valued sequences, such that each sequence only has finitely many non-zero entries. Consider the metric functions $d_2(\vec a, \vec b) = \sqrt{\sum_{n=1}^\infty |a_n - b_n|^2}$, and $d_\infty(\vec a, \vec b) = \max\{|a_n - b_n|, n=1,2,\cdots \}$. Prove that the two metric defines different topology on $P$. Hint: prove that there does not exists $c_1, c_2>0$, such that for all $\vec a \in P$, we have
+$$ c_1 d_2(0, \vec a) < d_\infty(0, \vec a) < c_2 d_2 (0, \vec a). $$
+Suppose there exists $c_1,c_2>0$ with the above property. Consider $\vec a_N = (1, \cdots, 1, 0, \cdots 0)$ where there are $N$ many $1$. Then $d_\infty(0, \vec a_N)=1$, $d_2(0, \vec a_N) = \sqrt{N}$. It is impossible that $c_1 \sqrt{N} \leq 1$ for all positive integer $N$, hence we have a contradiction.

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