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10

The distance between two sets of the same metric space is defined as: $$d(A,B) = \inf_{x\in A,\ y\in B} d(x,y)$$ That means that if $x\in A$ and $y\in B$ then $d(x,y) \geq d(A,B)$. Now, $d(x,y)$ is always finite in a metric space so $d(A,B)$ must be too.


5

If $X$ is a compact metric space, then $X$ itself is a compact subset of $X$, so the property $A$ must hold for $X$. If $X$ is not compact, then let $A$ be the property that a subset of $X$ is compact. Then $A$ holds for any compact subset of $X$, but $A$ does not hold for $X$ itself.


4

Let $x$ be an element of $A$ and let $y$ be an element of $B$. We know that $\|x-y\|$ is a real number $r$ and so $d(A,B)$ must be at most $r$ by the definition of $d(A,B)$, hence $d(A,B)$ is finite.


3

The whole space $X$ is closed, so the answer is no. Now let us change the problem to ask whether it is possible for every closed set apart from $X$ to be finite. Let $p$ and $q$ be distinct points. Let $\epsilon=d(p,q)/4$. Let $B_p$ be the open ball with centre $p$ and radius $\epsilon$, and define $B_q$ similarly. If every closed set apart from $X$ is ...


2

Each of your finitely many $U_i$ is a neighbourhood of $x$, hence by definiiton contains an open ball $B_{r_i}(x)$ for some positive number $r_i$. Pick one for each $i$. Let $r$ be the minimum of these finitely many positive real numbers, so $r$ is itself a positive real number. Consider $B_{r/2}(x)$. As $B_r(x)\setminus B_{r/2}(x)$ is nonempty, we have thus ...


2

1. Metrics that satisfy $d(x+r,y+r)=d(x,y)$: There are many metrics of this kind, as remarked by other users. Also the sum of any such metrics still has the property. 2. Metrics that satisfy $d(xr,yr)=d(x,y)$: There is such a metric on $(0,\infty)$, namely $d(x,y)=|\log(x)-\log(y)|$. There are also such metrics on $\mathbb R$, and something can be said ...


2

The characterization of the topology on $M$ that you have written (which, I see, is the characterization on the wikipedia page for pseudometrics) has an error. It should be: "a set $A$ in $M$ is closed (open) iff $A$ is saturated and $h(A)$ is closed (open) in $M^*$".


2

An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $\|Tx-Ty\|=\|x-y\|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $\|Tx\|=\|x\|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of ...


1

The uniqueness of $(Y,d_Y)$ in your modified version of the theorem is certainly not true. Precisely the point of saying that it is unique up to an isometry is that instead of $Y$ you can take any set $Y'$ such that there exists a bijection $Y\to Y'$. (And you transfer the metric form $Y$ to $Y'$ using this bijection; i.e., you take the metric on $Y'$ such ...


1

How about $d(x,y)=\max\{1,|y-x|\}$ or $d(x,y)=\frac{|y-x|}{1+|y-x|}$?


1

If the space consists of binary $m\times n$ matrices for fixed integers $m,n \gt 0$, then the number of such matrices is finite. Any finite metric space has a discrete topology, so there is only one metric topology up to topological equivalence. Indeed strong equivalence of metrics must also hold in this finite setting. Let $d_1(x,y)$ and $d_2(x,y)$ be two ...


1

Assume there exists precisely one geodesic segment from $p$ to $q$, say $\gamma : [0,l] \to M$, parametrized by arc length. By compactness we can extend $\gamma$ to the intervall $[0,\infty[$. Now consider a sequence $\gamma_n : [0,l_n] \to M$ of minimal arc length geodesics from $p$ to $\gamma (l + \frac 1 n)$. Since limits of minimal geodesics are minimal ...


1

Thanks wspin. However, let me elaborate a bit, some detail is still not clear to me. For any $s \in (-\epsilon,\epsilon)$, let $\gamma_s(t)$ be a minimal geodesic connecting $\gamma(0)$ with $\gamma(t+s)$. Let $V(t) :=\partial_s \gamma_s(t)|_{s=0}$ be the associated vector field along the original geodesic $\gamma$. By construction $V(0) = 0$ and $V(1) = ...


1

If $N_r(x)=\{y\in X:d(x,y)<r\}$ is the neighborhood of $x$ of radius $r$, then given $x\neq y$, let $r=\frac{d(x,y)}{2}$. Then $N_r(x)\cap N_r(y)=\emptyset$. But $N_r(x)$ and $N_r(y)$ are non-empty and open, so they must be the complements of closed sets. If only finite sets are closed sets, can $N_r(x)^c\cup N_r(y)^c=X$?


1

If $X$ is a countably infinite set, then there exists a bijection $f:X\to \Bbb Q$. Now what about the metric $d(x,y)=|f(x)-f(y)|$.


1

A vector space can be associated with different norms, all satisfying norm axioms. In your example, the first is $L^\infty$ norm and the second $L^1$ norm.


1

If $d$ is a metric, then $\lambda\cdot d$ is an equivalent metric to $d$ for any $\lambda\in\mathbf R$. Assume that $X=\{x_1,\dots,x_n\}$. If $d$ is a metric on $X$, define $c_{i,j}:=d(x_i,x_j)$. The set of all possible metrics on $X$ is equipotent to a subset of $\mathbf R^{n^2}$, which is equipotent to $\mathbf R$.


1

Let $(X,d)$ a metric space and $(x_n)_n$ a Cauchy sequence, and $a\in X$. $\epsilon=1$. There is a $N$ such that $\forall m,n\ge N$ $$d(x_m,x_n)<1$$ by triangular inequality we have $d(x_n,a)\leq d(x_n,x_N)+d(x_N,a)$, then for $n\geq N$ we obtain $d(x_n,a)\leq 1+d(x_N,a)=C$. Then If we take $M=\max\{d(x_0,a),\dots,d(x_{N-1},a),C\}$. Then $d(x_n,a)\le ...


1

Let $X$ be an infinite set. Note that a topology $\tau$ on X is a collection of subsets of $X$, that is $\tau\subset P(X)$. We consider the collection $T$ of all topologies on X. Pick a topology $\tau$ in $T$ one at a time, and let $(X,\tau)$ be the corresponding topological space. If $(X,\tau)$ is a metrizable space, then let a corresponding metric ...



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