Questions tagged [metric-spaces]

Metric spaces are sets on which a metric is defined. A metric is a generalization of the concept of "distance" in the Euclidean sense. Metric spaces arise as a special case of the more general notion of a topological space. For questions about Riemannian metrics use the tag (riemannian-geometry) instead.

A function $d: M\times M\to \mathbb R$ is called a metric if for all $x,y,z \in M$ we have

  1. $d(x,y)=0\iff x=y$
  2. $d(x,y)\geq 0$
  3. $d(x,y)=d(y,x)$
  4. $d(x,y)+d(y,z)\geq d(x,z)$.

It is a generalisation of "distance". A metric space is now defined as an ordered pair $(M,d)$, where $M$ is a set and $d:M\times M\to R$ is a metric.

An $\varepsilon$-neighbourhood of $x$ is defined as the set $$B_\epsilon(x):=\{y\in M\mid d(x,y)<\varepsilon\}.$$ $B_\varepsilon(x)$ is commonly also known as the open ball of radius $\varepsilon$ around $x$. All open balls form a base for a topology on $M$. Although all metric spaces are topological spaces, the converse is generally not true.

Some different types of metric space include

  1. Complete metric spaces (every Cauchy sequence converges)

  2. Bounded metric spaces (every metric is bounded by a finite value)

  3. Compact metric spaces (every sequence has a convergent subsequence)

  4. Locally compact metric spaces (every point has a compact neighbourhood)

  5. Separable metric spaces (it possesses a countable dense subset).

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Since every subset is open, every subset is also closed

The answer to the question is given below: Could somebody please explain me how do they deduce that "Since every subset is open, every subset is also closed". Is this some kind of definition, property? The way I see that the sets are both open…
user634512
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why $H^\mathrm{o}=K^\mathrm{o}=\emptyset$

$A^\mathrm{o}$, the interior of $A$, is the union of all open subsets of $A$. If $H=\Bbb Q$ and $K=\Bbb R \backslash \Bbb Q$ then $H^\mathrm{o}=K^\mathrm{o}=\emptyset$ but $(H\cup K)^\mathrm{o}=\Bbb R^\mathrm{o}=\Bbb R$ I don't understand why…
user634512
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Metric spaces, continuous functions

Let $f,g: M \rightarrow \mathbb{R}$ be continuous in $a \in M$. If $f(a)0$ such that, for all $x,y \in M$ $$d(x,a)<\delta, d(y,a)<\delta \rightarrow f(x)
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Confusion about a "scalable" metric

I'm reading a book called Geometry of quantum states. On page 18 the authors introduce the usual notion of metric on a vector space and continue to discuss Minkowski distance. A metric $d$ on a vector space $V$ is called a Minkowski distance…
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Show that limits of sequences are unique in metric spaces

Let $(\mathbb{X}, ρ)$ be a metric space. Let $\{x_k\}_{k=1}^{\infty}$ be a convergent sequence in $\mathbb{X}$ with a limit $\lambda \in \mathbb{X}$. Show that the limit $\lambda$ is unique. We know convergence means, $\forall \frac{\epsilon}{2} >…
user486957
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To prove metric space

Let $H$ be the set of all sequences of real numbers $x={X_n} $ such that $|X_n|\leq1 $ for all $n$ belonging to Natural Number set $\mathbb{N}.$ Consider the function $d: H X H \to R $ given by $d(x,y) = \sum[(|X_n-Y_n|)/Z^n] $, where $x={X_n} $ and…
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$A=\{x_n:n\in\mathbb N\}$ is a closed and discrete subset of $X$ w.r.t. the metric induced from $d.$

Let $(X,d)$ be a metric space and $(x_n)$ be a sequence in $X.$ If $(x_n)$ has no cluster point in $X$ then prove/disprove: $A=\{x_n:n\in\mathbb N\}$ is a closed and discrete subset of $X$ w.r.t. the metric induced from $d.$ I think the statement…
Jave
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If $(X,d_1)$ and $(Y,d_2)$ be metric spaces . Is there a metric on $X \cup Y$ which induces $d_1$ on $X$ and $d_2$ on $Y$?

If $(X,d_1)$ and $(Y,d_2)$ be metric spaces. Is there a metric on $X \cup Y$ which induces $d_1$ on $X$ and $d_2$ on $Y$? (we can assume $X \cap Y = \emptyset$).
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Diameter of open and closed balls

I'm homelearning calculus and metric spaces. We are in a space with euclidean metric. In my book, it is stated that the diameter of $S(1,2)=2$ and the diameter of $\bar{S}(1,2)=4$, where $S$ is an open ball and $\bar{S}$ is a closed ball. Could you…
LukasT
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Volume of a $l_2$ ball of radius $r$

What is the volume of a ball $B \subset \mathbb{R}^n$, where $B = \{ \theta : ||\theta||_2 \leq r\}$ where $|| \cdot ||_2$ is the $l_2$-norm? The $l_2$-norm is given by $||x||_2 := \left( \sum_{i=1}^d |x_i|^2 \right)^{1/2}$.
kkc
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Prove or disprove that $(0,1)$ complete.

A metric is defined on $X=(0,1)$ by $d(x,y)=\left|\frac 1x -\frac 1y\right|$. Is this space complete or not? My attempt: I have shown that there does not exist any $x_n$ that tends to 0 ($x_n\neq 0$ for all $n$). Because if $x_n$ tends to 0 then…
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Prove that $d(x,y)=\frac{1}{x+y}$ is not a metric space

I am supposed to prove that: $d(x,y)=\left\{\begin{matrix} 0 & x=y \\ \frac{1}{x+y} & x\neq y \end{matrix}\right.$ is not a metric space, $\forall x,y \in \mathbb{N}$ The first two properties for the metric space holds, also the triangle…
Shelley
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$\sup\limits_{y\in [0,1]\setminus\{x\}}\left|\frac{f(x)-f(y)}{x-y}\right|\in \mathbb{R}_+$ continuous?

Is this function $f\in(\mathcal{C}([0,1],\mathbb{R}),\|\cdot\|_\infty)\mapsto B(f)=\sup\limits_{y\in [0,1]\setminus\{x\}}\left|\dfrac{f(x)-f(y)}{x-y}\right|\in \mathbb{R}_+$ continuous? I think if I reduce this metric space…
Stu
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Equality of diameters in metric space

Does it hold that in any metric space: $diam S\left ( p,r \right )= diam \tilde{S}\left ( p,r \right )$? My solution: When we consider trivial metric, where: d(x,y)=0, if x=y d(x,y)=1, if x $\neq $ y, then it does, because diameter is the…
Martin N.
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Identical metric spaces

My class was deleted, due to the virus. I am supposed to solve the following problem: Let $A$ be any set of positive real numbers. Construct a metric space $(X, d)$ so that the set $A$ is identical to the set of all distances of different points of…
Shelley
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