Show that $(\Bbb N,d_1)$ and $(\Bbb N,d_2)$ are equivalent where $d_1$ is the discrete metric and $d_3(m,n)=|\frac{1}{m}-\frac{1}{n}|$.   My account  In order to show that $(\Bbb N,d_1)$ and $(\Bbb N,d_2)$ are equivalent,take any $x\in \Bbb N$ and consider $B_{d_1}(x,r)$ .We need to find $s&gt;0$ such that $x\in B_{d_3}(x,s)\subset B_{d_1}(x,r)$ and vice versa.  If we choose $s=r$ in the first case then if $y\in B_{d_1}(x,s)\implies d_1(x,y)&lt;s\implies |x-y|&lt;s\implies |\frac{1}{x}-\frac{1}{y}|&lt;s\implies y\in B_{d_3}(x,r)$.  But I am stuck in the other subset inequality.Please help .  Is there any other way to solve this ?
120925
How to show that $\mathbb N$ when given the metric $d(m,n)=\dfrac{1}{m}-\dfrac{1}{n}$ and when given the subspace topology as inherited from $\mathbb R$ are equivalent.  When $\mathbb N$ is given the subspace topology then $\mathbb N$ becomes the discrete space.But I cant show that in the first case single point sets are open.  I know to show equivalent we have to proceed as   take any $x\in B(x,r)$ and then find $s&gt;0$ such that $x\in B(x,s)\subset B(x,r)$  But I cant find the $s&gt;0$ in the metric given .Please help
69557
Suppose we have two equivalent norms, $\|\cdot \| \sim \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert$. We assume that $x_n \rightarrow x $ in $(X,\|\cdot \|)$. Show that $x_n \rightarrow x $ in $(X,\left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert)$  We observe since the norms are equivalent, there exits constants $\alpha, \beta$, such that $\alpha \|\cdot \| \leq \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert \leq \beta \|\cdot \|$, with $0 \lt \alpha \leq \beta$  Since, $x_n \rightarrow x$ in $(X,\|\cdot \|$) we know that for all $\epsilon&gt;0, \exists K(\epsilon)$ such that for n>$K(\epsilon),$ we have $\|x_n -x\| \leq \epsilon$  Since, we know $\beta \gt 0$ , then $ \beta \|x_n -x\| \leq \beta \epsilon$, Thus we see that $\left\vert\!\left\vert\!\left\vert x_n-x\right\vert\!\right\vert\!\right\vert\leq \beta \epsilon$.  This is how far I've gotten, our goal to is to find a a $J(\epsilon)$ such that for all $\epsilon&gt;0, \exists J(\epsilon)$ such that for n>$J(\epsilon),$ we have $\left\vert\!\left\vert\!\left\vert x_n-x \right\vert\!\right\vert\!\right\vert \leq \epsilon$  My candidate for $J(\epsilon$) is $J(\epsilon) = \frac{K(\epsilon)}{\beta}$ from my above argument, but I'm not exactly sure how to validate this. If someone could help me on the proof, it would be very appreciated. I believe I'm pretty close, just missing something small.
58380
Problem:     If $(k,l) = 1$, show that $(b_2k-b_1l, a_1l-a_2k) = 1$ for $a_1b_2-a_2b_1=1$   Note: ($a_1, a_2,b_1,b_2,k,l \in \mathbb{Z}$)   Also note that the actual (bigger) problem is:     If $m = a_1x+b_1y$ (Eq1) and $n=a_2x+b_2y$ (Eq2) and $a_1b_2-a_2b_1=1$, prove that $(m,n)=(x,y)$.   I have proved that $(m,n) = d | x \text{ and } d|y$ I am left to prove that $d = (x,y)$, that's why I asked above question.  I had made $m = dk$ and $n = dl$ ($\therefore (k,l)=1$) and after substitution in Eq1 and Eq2 for $x$ and $y$ values I got $d(b_2k-b_1l)=x$ and $d(a_1l-a_2k) = y$. If I can show  $(b_2k-b_1l, a_1l-a_2k) = 1$ then definitely $d = (x,y)$  Method:  Suppose $(b_2k-b_1l, a_1l-a_2k) = d &gt; 1$. Then, $d | b_2k, d | b_1l, d|a_1l,d|a_2k$ must be true.  This implies $d|a_2b_1kl$ and $d|a_1b_2kl$ must be true. Let, $dp = a_2b_1kl \implies a_2b_1 = \frac{dp}{kl}$ and $dq=a_1b_2kl\implies a_1b_2 = \frac{dq}{kl}$  So, $$a_1b_2-a_2b_1=1 \implies \frac{d(q-p)}{kl}=1 \implies kl =d(q-p)$$  I am stuck here. How to prove a contradiction so that $d = 1$?
17569
Let $(X,d)$ be a metric space and $(a_n)$ a Cauchy sequence in $X$. How to show that there exists $n_1&lt;n_2&lt;\cdots$ such that $$d(a_{n_{i+1}},a_{n_i})&lt;2^{-i},$$ for all $i\in\mathbb{N}$?  I tried to use induction, but it doesn't work. The base case is OK, but if we have $$d(a_{n_2},a_{n_1})&lt;2^{-1},\ldots,d(a_{n_k},a_{n_{k-1}})&lt;2^{-(k-1)},$$ I don't see any reason for to exist $n_{k+1}&gt;n_k$ such that $d(a_{n_{k+1}},a_{n_k})&lt;2^{-k}$.
53976
I'm new to topology, and can't figure out why the metric and product topologies over $\mathbb{R}^n$ are equivalent.  Could someone please show me how to prove this?
68258
Let $S$ be the set of all real sequences , for $(x_n) , (y_n) \in S$ , let $d \Big((x_n) , (y_n) \Big) :=0 $ if $x_n=y_n , \forall n \in \mathbb Z^+$ otherwise $d \Big((x_n) , (y_n) \Big) :=\dfrac 1{\min \{n :x_n \ne y_n\}}$ , then is $d$ a metric ?   EDIT : I have only difficulty in proving the triangle inequality
29374
Let $f\in H(B(0,1))$, $f(0)=0$, $f(B(0,1))\subset B(0,1)$. Suppose there are 2 different points $z_1$,$z_2\in B(0,1)$, s.t. $|z_1|=|z_2|$, $f(z_1)=f(z_2)$. Show that $|f(z_1)|=|f(z_2)|≤|z_1|^2=|z_2|^2$.    I have used the Schwarz-Pick Proposition to prove that $|f(z_1)|=|f(z_2)|≤|z_1|=|z_2|$ directly, but that is not enough. Thank you for help!
65335
How would I show that $P(\mathbb{N})$ is equivalent to $2^\mathbb{N}$? The questions asks to form a theorem from the statement I just gave. Any help would be appreciated.
20957
I tried $|d(x, z) - d(y, z)| \leq d(x, z) + d(y, z)$, but the the triangle equation goes the wrong side. Is there an analogue to the reverse triangle inequality with metrics? The reason I want to know this is to proof that $d(x, A)$ is continious where $A$ is a set. (Let $|x - y| &lt; \delta$, then after some algebra $|d(x, A) - d(y, A)| = \inf(\{|d(x, a) - d(y, a)| \ | \ a \in A \})$.
83108
I am trying to work just one case of showing that the French railways metric defined by a metric space $(\mathbb{R}^2,d)$ is actually a metric:  $$d(x,y) = \begin{cases} \|x-y\|, &amp; \text{if $x,y,0$ are collinear;} \\ \|x\| +\|y\|, &amp; \text{otherwise} \end{cases}$$  I am trying to show the case that if $x=y$, then $d(x,y)=0$. I have been able to show that if $x=y$, then $\|x-y\|=0$, but am having a hard time showing that if $x=y$, then  $\|x\| +\|y\|=0$. Thank you!
10084
I'm currently stuck on an exercise where I need to show that if $(M,d)$ and $(N,\rho)$ are metric spaces, then $d_2((a,x),(b,y)) := \sqrt{d(a,b)^2 + \rho(x,y)^2}$ defines a metric on $M\times N$. I've already shown that the basic criteria of a metric are met, but I still need to show that the triangle ineqaulity holds.  I've tried using the triangle inequality on $d$ and $\rho$ and estimating the square root with $\sqrt{a^2+b^2}\leq a+b$, but regardless of which order I've used them in, I seem to overestimate my target. I thought maybe I somehow need to use Cauchy-Schwarz, but I don't know how.
97688
Hi I just want to find out if the following is a acceptable proof for the proposition:   "Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."  Proof:  Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed connected, either $A \cap V_{1} = \emptyset$,  $A \cap V_{2} = \emptyset$, $A \cap V_{1} = A$ or $A \cap V_{2} = A$.   W.L.O.G assume that $A \cap V_{1} = \emptyset$, then there exists an $x \in B \subset \bar{A}$ and $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A = \emptyset$. This contradicts $x \in \bar{A}$. Assume now W.L.O.G that $A \cap V_{1} = A$, then $A \subset V_{1}$ and $A \cap V_{2} = \emptyset$. If we take $x \in V_{2} \subset B \subset \bar{A}$ then again we find $\epsilon &gt; 0$ such that $B(x,\epsilon) \cap A =\emptyset$. This contradicts $x \in \bar{A}$. Therefore there cannot exist disjoint open sets $V_{1}$ and $V_{2}$ in $B$ such that $B = V_{1} \cup V_{2}$. Hence, $(B,d)$ cannot be disconnected and is therefore connected.   $\square$
22190
So far I have that $a|b$ implies $b=ax$ for some x in the integers and $c|d$ implies that $d=cy$ for some y in the integers.  From here I can see that $gcd(a,c)|gcd(b,d)$ is logically equivalent to $gcd(a,c)|gcd(ax,cy)$ but I am not sure where to go from there.
85677
Where $ (X, d) $ is a metric space. I want to prove it using sequential criteria.  How do I tackle it?
106915
If $\mathcal{M}=\{M_i : i\in I_n\}$ is a collection of metric spaces, each with metric $d_i$, we can make $M=\prod_{i\in I_n}M_i$ a metric space using the $p$-norm, we simply set $d : M\times M\to \mathbb{R}$ as:  $$d((p_1,\dots,p_n),(q_1,\dots,q_n))=\left\|(d(p_1,q_1),\dots,d(p_n,q_n))\right\|_p$$  What I want to prove is that the $p$-norm  $$\left\|x\right\|_p=\left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}$$  is really a norm. Showing that $\left\|x\right\|_p \geq 0$ being zero if and only if $x = 0$ was easy. Showing that $\left\|kx\right\|_p = \left|k\right|\left\|x\right\|_p$ was also easy. The triangle inequality is the thing that is not being easy to show. Indeed, I want to show that: for every $x,y \in \mathbb{R}^n$ we have:  $$\left(\sum_{i=1}^{n}\left|x_i+y_i\right|^p\right)^{1/p}\leq \left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}+\left(\sum_{i=1}^{n}\left|y_i\right|^p\right)^{1/p}.$$  I thought that it might not be as difficult as it seems, but after trying a little without sucess I've searched on the internet and the I found that we need measure theory to prove that. Is there any more elementary proof of this inequality?
65614
I am trying to understand the following statement from an analysis book:     Two norms are equivalent ($||\cdot ||_2 \sim||\cdot||_1 $) if they   induce equivalent metrics.   At first I thought this was straight forward but then the book has the following proposition that leads me to believe that I have a misunderstanding:     Let $||\cdot||_1$ and $||\cdot||_2$ be two norms on a field F. Then   $||\cdot ||_2 \sim||\cdot||_1$ iff there exists a positive real number   $\alpha$ such that $||x||_2=||x||_1^\alpha , \ \forall x\in F$.   How can it be that $||x||_2=||x||_1^\alpha$ if x is any other value other that 1 if $\alpha\ne$1? Maybe this is related to my misunderstanding... can norms and metrics be thought of as the same thing? Thanks!
20046
Suppose $A, B, C$ are categories. If $A$ and $B$ are equivalent, is it the case that $C^A$ and $C^B$ are equivalent? Also, is it the case that $A^C$ and $B^C$ are equivalent.  I first conjectured that $C^A$ and $C^B$ are equivalent, but then as I tried to prove the conjecture, I started to realize that it might not be the case. So now I'm trying to find a counterexample. So far I have not been able to find one. Maybe I'm stuck with "too complicated" ideas. I strongly believe that there are very simple counterexamples. Please help. (For more detail, I got to the point where I have a cube that I want to be commutative, but I can only show that 4 of its sides are commutative squares. That's when I realize the other two sides may not necessarily commute.)  And of course, there is the second part to the problem that's not just the dual problem. I would like to know about the special case in which $A$ is a preorder and $B$ is a partial order which is the skeleton of $A$. This may have nothing to do with the validity of the statement in question at all, but it is the setting that originated this question.