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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Are $d_{\infty}$ and $d_{p}$ distance functions?

Let $X$ be a set equipped with a metric $d_x$, denoted by $\langle X,d_x\rangle$, and $Y$ equipped with a metric $d_y$, denoted by $\langle X,d_y\rangle$. Let $Z=X\times Y$. Let $z_1=(x_1,y_1), \ z_2=(x_2,y_2), \ \forall z_1,z_2\in Z$, we define…
Ingvar
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Enlargement or reduction of metric may destroy completeness

I'm looking for examples in which the following occurs: We have a set $M$ and metrics $d_1$, $d_2$ on $M$ such that $\forall x, y \in M: d_1(x, y) \leq d_2(x, y)$. And then either: $(M, d_1)$ is complete but $(M, d_2)$ is not $(M, d_2)$ is complete…
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Is there any open set in $(\mathbb R,d)$ which is not open in $(\mathbb R,d')$ where $\mathbb R$ is set of real number?

Here $(\mathbb{R},d)$ is metric space with usual distance metric and $(\mathbb{R},d')$ is metric with discrete metric. I think there is no this type of set in $(\mathbb{R},d)$ which is not open in $(\mathbb{R},d')$.
ëlêtro
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Coordinate-Additive Metric

Let $R$ be the real numbers. I want to consider metrics $d: R^n \times R^n \rightarrow R$ such that there are functions $h: R \rightarrow R$ and $\delta: R \times R \rightarrow R$ with $$ d(x,y) = h \left(\sum_{i=1}^n \delta(x_i,y_i)…
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Every $2^{−k} r$ -separated set in every ball $B(x , r )$ in $X$ has at most $N^k$ points in a doubling space

Let $X$ be a doubling metric space with constant $N$ and let $k \geq 1$ be an integer. Then every $2^{−k} r$ -separated set in every ball $B(x , r )$ in $X$ has at most $N^k$ points. Let $S$ be a maximal $2^{-k}$ separated subset of a $B(x,r)$…
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Let $(\Omega, d)$ be a metric space. Prove that the sequence $x_n=1 / n, n \in \mathbb{N}$ does not converge if $\Omega=(0,1], d(x, y)=|x-y|$.

Let $(\Omega, d)$ be a metric space. A sequence $\left(x_n\right)_{n \in \mathbb{N}}$ in $\Omega$ converges to an element $x \in \Omega$ if $d\left(x_n, x\right) \rightarrow 0$ as $n \rightarrow \infty$, i.e., $$ \forall \epsilon>0 \exists N \in…
Mark
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If $A$ is everywhere dense in the metric space $M$, then the only closed set which contains $A$ is $M$.

If $A$ is everywhere dense in the metric space $M$, then the only closed set which contains $A$ is $M$. My attempt: Let $B$ be a closed set which contains $A$. My aim: Prove $B=M$ Since $A$ is everywhere dense in $M$, $\bar{A}=M$. By the theorem…
Idonknow
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$A$ is a set of points in ametric space $M$ and $B$ is the set of all accumulations points of $A$. Prove that $B$ is closed.

$A$ is a set of points in a metric space $M$ and $B$ is the set of all accumulations points of $A$. Prove that $B$ is closed. Aim: Prove that $B^{\complement}$ is open. Let $y \in B^{\complement}$. Then $y$ is not an accumulation point of $A$. Then…
Idonknow
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Proving a particular subset of $R^n$ is closed

Let $S,X$ be subsets of $R^n$ given by $$S=\{(a_1,a_2,\dotsc,a_n)\in R^n|\sum a_i^2=1\}$$ $$X=\{(b_1,b_2,\dotsc,b_n)\in R^n|\sum\frac{b_i}{i}=0\}$$ Then prove that $S+X$ is a closed set in $R^n$.
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Confusion regarding totally bounded metric spaces having cauchy subsequences of every sequence.

My book says "A metric space $(X,d)$ is totally bounded if and only if every sequence in $X$ contains a cauchy subsequence." Let us take a sequence $\{x_i\}$ such that for every prime number $p$, when $i$ is of the form $p^k$, $x_i$ or $x_{p^k}$ is…
user67803
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Proving a set is not an open ball

Let $(X,d)$ be a metric space, $x\in X$ and $r>0$. The set $$B(x,r)=\{y\in X: d(x,y)
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Compactness of $(a,b)$ in $\Bbb{R}$.

Say we have an open set $(a,b)\subset\Bbb{R}$, which has an infinite cover $\mathfrak{C}$. Let us assume $(a,b)$ is not compact. Then we can select an open set $(a_1,b_1)\subset (a,b)$ such that it is not compact. If we cannot select such an…
user67803
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Cantor's intersection theorem for nested open sets.

Let $\{I_k\}$ be a set of nested closed bounded sets such that $\forall k\in\Bbb{N}, I_k\supset I_{k+1}$, and $\lim_{n\to\infty}\operatorname{diam}(I_n)=0$. Let the metric space $(X,d)$ be complete. This implies that $\bigcap_{k=1}^\infty…
user67803
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Investigating the importance of boundedness in Heine-Borel's theorem.

We know that $[1,\infty)$ is not compact in $\Bbb{R}$. A simple proof: the cover $\mathfrak{C}=\bigcup B(n,\frac{3}{4})$ for all $n\in\Bbb{N}$ of $[1,\infty)$ does not have a finite subcover. What is wrong with the following argument (which closely…
user67803
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Metric space with an uncountable subset containing points separated by distance greater than one

Let $\langle X,d \rangle$ be a metric space such that there is an uncountable set $Y \subseteq X$ such that any two distinct points of $Y$ have distance greater than one. Show that $X$ is not separable. I know this is a very similar problem, but…