Show that $(\mathbb{N}, d)$ is a metric space   with $$d(x,y) = \frac{|x-y|}{1+|x-y|}$$   My attempt:   let $x,y \in \mathbb{N}$,  1) $d(x,y) = 0 \implies |x-y| = 0 \iff x = y$  2) $d(x,y) = d(y,x)$  3) show for $z \in \mathbb{N}$, $d(x,z) \leq d(x,y) + d(y,z)$  Using the triangle ineuqality I managed to get:  $$d(x,z) \leq \frac{|x-y|}{1+|x-y| - |y-z|} + \frac{|y - z|}{1 + |x-y| - |y-z|}$$  but i assume here I must use the fact that $x,y,z \in \mathbb{N}$ but I'm not sure how.  Any hints please:  edit: improvements on solution - Using fact that $|x-y|$ is a metric we have that $$d(x,y) \leq \frac{|x-y|}{1+|x-y| + |y-z|} + \frac{|y - z|}{1 + |x-y| + |y-z|}$$ (Since $|x-z| \leq |x-y| + |y-z|$) Is this correct? From there I can use the fact that $$\frac{1}{1+|x-y|+|y-z|} \leq \frac{1}{1+|x-y|} $$ and $$\frac{1}{1+|x-y|+|y-z|} \leq \frac{1}{1+|y-z|}$$ to conclude?
121436
Show $\rho (x,y)=\dfrac{d(x,y)}{1+d(x,y)}$ is a metric on the metric space $X$, equipped with the Euclidean metric $d$.   I've already shown that the positivity $\rho(x,y)\geq 0$, the symmetry $\rho(x,x)=\rho(y,y)=0$.  I'm having trouble proving the last condition of a metric space though: the triangle inequality must be valid:  $\rho(x,y)\leq \rho(x,p)+\rho (p,y)$, for $p\in X$. So, $\dfrac{d(x,y)}{1+d(x,y)}\leq \dfrac{d(p,x)}{1+d(p,x)}+\dfrac{d(p,y)}{1+d(p,y)}$  Could someone provide a hint as to get started? No complete solutions please. Thanks in advance.
99348
Say I have a metric space $(X, d)$ and a set $A\in X$.  I want to prove that if $a \in \overline{A}$ then there is a convergent sequence $\{x_n\}$ that converges to $a$. Could I have any help?
79542
Let $A,B \subseteq \mathbb{R}^n$ with euclidean metric. Furthermore let $$ \mbox{dist}(X, Y) := \inf\{|x - y| : x \in X, y \in Y\}. $$ Does the following implication hold?     $A \subseteq B \implies \forall x \in A: \mbox{dist}(x,\partial A) \le \mbox{dist}(x, \partial B)$    Problems could arrise if one of the boundaries $\partial A$, $\partial B$ is empty. But in this case $A$ or $B$ would be clopen, and hence equal to $\mathbb{R}^n$ or $\emptyset$. I think I am better off assuming $A \neq \emptyset$ and $B \neq \mathbb{R}^n$ to avoid problems.    Thanks to a suggestion by GEdgar I came up with the following proof:  Let $s \colon [0,1] \to \mathbb{R}^n$ be the straight line with $s(0) = x \in A$ and $s(1) = y \in \partial B$. I want to show, there exists $t' \in [0,1]$ with $s(t') \in \partial A$.   It is safe to assume $x \not \in \partial A$, since then $\mbox{dist}(x,\partial A) = 0$ and the inequality holds trivially. Same goes for the assumption $y \not \in \partial A$.  Consider now the function $$ f(x) := (1-2 \chi_A(x)) \cdot \mbox{dist}(x,\partial A). $$ One can check that this function is continuous on $\mathbb{R}^n$:   No problems on $\mathbb{R}^n \setminus \partial A$, since as a property of metric spaces $\mbox{dist}(\cdot,\partial A)$ is continuous and furthermore $(1-2 \chi_A(x))$ is constant. Let $\lim_{n \to \infty} x_n = x \in \partial A$. Then  $$ |(1-2 \chi_A(x_n)) \cdot \mbox{dist}(x_n,\partial A)| \leq \mbox{dist}(x_n,\partial A) \to 0 = \mbox{dist}(x,\partial A), $$ since the closedness of $\partial A$ implies $\mbox{dist}(x,\partial A) = 0$ iff $x \in \partial A$.   Now the claim follows by application of the intermediate value theorem.  Is this proof correct?  Does the claim also hold for more general metric spaces $(X,d)$ which are only connected/path connected? Of course the definition of $\mbox{dist}$ would be modified accordingly $(| x - y | \to d(x,y))$
23792
Let $X$ denote a metric space.  Whenever $S \subseteq X$ and $r \in \mathbb{R}_{\geq 0}$, write $f(S,r)$ for the following set.  $$\{x \in X \mid \exists s \in S : d(x,s) \leq r\}$$     Question. Given a closed set $S \subseteq X$ and any $r \in \mathbb{R}_{\geq 0}$, does it follow that $f(S,r)$ is closed?   If not, a counterexample $(X,S,r)$ would be appreciated.
40609
Prove that a closed ball in a metric space is a closed set   My attempt: Suppose $D(x_0, r)$ is a closed ball. We show that $X \setminus D $ is open. In other words, we need to find an open ball contained in $X \setminus D$.   Pick $$t \in X-D \implies d(t,x_0) &gt; r \implies d(t,x_0) - r &gt; 0 $$ Let $B(y, r_1)$ be an open ball, and pick $z \in B(y,r_1)$. Then, we must have $d(y,z) &lt; r_1 $. We need to choose $r_1$ so that $d(z,x_0) &gt; r$. Notice by the triangle inequality  $$ d(x_0,t) \leq d(x_0,z) + d(z,t) \implies d(z,x_0) \geq d(x_0,t) - d(z,t) &gt; d(x_0,t) - r_1.$$  Notice, if we pick $r_1 = d(t,x_0)-r$ then we are done.  Is this correct?
106709
Let $(M,d)$ be a metric space. For a subset $A\subseteq M$ we define the distance of a point $x$ to $A$ as $$\alpha_A(x):=\operatorname{dist}(x,A):=\inf_{y\in A}d(x,y)$$ Prove that: a) $|\alpha_A(x_1)-\alpha_A(x_2)|\leq d(x_1,x_2)\text{ for all }x_1,x_2\in M$. b) $\alpha_A:M\to\mathbb{R}$ is continuous. c) $\alpha_A(x):=0\leftrightarrow x\in\bar{A}$. d) If $A$ is closed and $K\subseteq M$ compact with $A\cap K=\emptyset$, then $A$ and $K$ have a positive distance, meaning $$\operatorname{dist}(K,A):=\inf d(x,y)&gt;0\text{ for }x\in K,y\in A.$$  My ideas so far: a) I can imagine that this is true because I imagine $A$ being a ball and individual distance on $\alpha$ is the closest distance to that ball. But I don't know how to write down the proof mathematically. b) Now I'm lost here. Don't know how to approach this. c) Again, I can easily imagine that since $x$ is in $A$, meaning the closest distance to the subset is $0$, but I can't prove that mathematically. d) Similar to before, I have no idea how to write down the mathematical proof. But I imagine this one as a situation with two separate balls, since the intersection is the empty set. Meaning that the &quot;gap&quot; between them is the positive distance. Can someone show me how to write down mathematical proofs on this type of questions?
47817
Let $(M,d)$ be a compact metric space and $f:M \to M$ such that $d(f(x),f(y)) \ge d(x,y) , \forall x,y \in M$ ; then how to prove that $d(f(x),f(y))=d(x,y) , \forall x,y \in M$ i.e. that $f$ is an isometry  ? I wanted to prove by contradiction , that suppose $\exists a,b \in M$ such that $d(f(a),f(b)) &gt; d(a,b)$ ; I wanted to use the sequences $\{f^n(a)\}$ and $\{f^n(b)\}$ , these have convergent subsequences $\{f^{r_n}(a) \}$ and $\{f^{r_n}(b)\}$ ( without loss of generality we have extracted a common subsequence ) converging to say $x,y \in M $ respectively . Then $d (f^{r_n}(a) , f^{r_n}(b))$ converges to $d(x,y) &gt;d(a,b)$ ;  but I don't know what to do next , please help . Thanks in advance
90315
Let $X$ be a metric space , $d$ is the metric , show that $d$ is a continuous function from $X\times X$ to $R$.  I think the definition is all we need , but I just don't know where to start , can anyone help me.
96662
I've been reading about metric spaces recently and had a question about the definition. Everywhere I've seen, we take a metric space to be a set $E$ equipped with a metric, a binary map $d:E\times E\mapsto \mathbb{R}$, where $\forall x, y, z\in\mathbb{R}$,  $d$ satisfies:   $d(x, y) \geq 0$ $d(x, y) = 0 \iff x = y$ $d(x, y) = d(y, x)$ $d(x, z) \leq d(x, y) + d(y, z)$   It isn't difficult to see that axioms $2$, $3$, and $4$ imply $1$, so $1$ doesn't add anything to the definition and is superfluous. Why then do we keep 1 as an axiom in the definition of a metric instead of it being a corollary to the definition?
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.
97024
Let $(X,d)$ be a metric space.  Prove:   the union of any open sets in $X$ is open in $X$ the intersection of a finite number of open sets in $X$ is open in $X$   I could prove the first one but how do we prove the second one?
18101
Let $(X, d)$ be a metric space. I have to prove the following statements are equivalent.   $(X,d)$ is complete (i.e., every Cauchy sequence is convergent) and totally bounded (i.e., for every $\epsilon&gt;0$, $(X,d)$ has a finite $\epsilon$-net). $(X,d)$ is sequentially compact (i.e., every sequence has a convergent subsequence) . $(X,d)$ is Bolzano - Weierstrass compact (i.e., every infinite subset of $X$ has a limit point). $(X,d)$ is compact (i.e., every open cover of $X$ has a finite subcover). Every infinite open cover of $(X,d)$ has a proper subcover.   I have proved the following: 1$\Longleftrightarrow$2, 2$\Longleftrightarrow$3, 4$\implies$3, 2$\implies$4, 4$\implies$5 and 5$\implies$3. Although they are enough to prove the equivalence of all five statements, I tried to prove it from the other possible directions. But I'm stuck with the proof of 3$\implies$4 and 5$\implies$4. Does anybody know how to prove them without using any of the above statements? Please help.
110759
I am given the following problem:     Show that if a metric space (X,d) is compact (meaning X is compact   with respect to the metric d), then there exist points a,b ∈ X such   that d(a,b) = sup{d(x,y) : x,y in X}. (Hint: You’ll need to use the fact   that (X,d) is compact twice. In particular, this means that every   compact metric space has an upper bound on how far away two points can   be)   However, I am not sure if the problem has left out $X \subseteq \mathbb{R}$. Does the statement hold in a general metric space?  Secondly, how might one go about proving this?
100239
In notes I have come across, it is stated that for a Hausdorff space induced by countably many semi-norms $p_n$ that,  $$d(x,y) = \sum_{n=1}^{\infty}2^{-n}\frac{p_n(x-y)}{1+p_n(x-y)}$$ is a metric.  But I can't see why this metric satisfies the triangle inequality. As I believe it satisfies the triangle inequality iff $p_n(x-y)(1+p_n(x-z)p_n(z-y)) \leq p_n(x-z) + p_n(z-y)$  If you shed some light on this, it would be much appreciated!
119812
In a metric space $(X,d)$, prove that for each subset $A$, $x\in\bar A$ if and only if $d(x,A)=0$  I feel like this isn't necessarily true. For example, let $X$ equal the reals and $A$ be some open subset of $X$, let's say $(0,1)$. Then $0.5$ is an element of $\bar{A}$, but $d(x,A) = d(0.5,0) \neq 0$  This question seems simple enough; I don't know why it's giving me such trouble.
12348
I have an exercise found on a list but I didn't know how to proceed. Please, any tips?  Let $X$ be a connected subset of a connected metric space $M$. Show that for each connected component $C$ of $M\setminus X$ that $M\setminus C$ is connected.
91073
Please ,how to prove that the space $\mathbb{R}$ endowed with the metric $d(x,y)=|e^x-e^y|$ is  not a complete space?  I don't find a Cauchy sequence   but not convergent   Please  Thank you.