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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Is every open set a disjoint union of open balls?

I know that in $\mathbb R$, every open set is a disjoint union of open intervals, i.e., the basic open balls. Is there a similar result that holds for arbitrary metric spaces? Suppose $X$ is a metric space and $U$ be any open set of $X$. Can I write…
user118494
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Hausdorff dimension

Could you please give me some hints on (Exercise 1.7.21) of "A Course in Metric Geometry" by Burago, Burago, Ivanov. We have a compact space $X$, which can be written as $X=\bigcup_{i=1}^n X_i$ (disjoint union), where $X_i=F_i(X)$ and $F_i$ are…
zerhan
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example of a metric space in which triangle inequality is equality

Is there any example of a metric space $X$ with more than two points such that the triangle inequality is always equality?
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A subspace $Y$ of a complete metric space $X$ is closed if it is complete.

For this problem, let there is a sequence $\{y_n\}$ which is complete in $Y$ such that this sequence has finitely many distinct points and this sequence approaches to say $y$. In this case $y$ is repeated infinitely many times. In this scenario,…
rohit
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How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$?

Let $f:(X, d_X)\longrightarrow (Y, d_Y)$ be a continuous function between two metric spaces, $A\subseteq X$. We have $f(\overline{A})\subseteq \overline{f(A)}$ from this question. Can you please provide a counter-example to $f(\overline{A})\supseteq…
ahorn
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Get a bounded metric from a metric - triangle inequality for $d'(x,y):=\frac{d(x,y)}{1+d(x,y)}$

This is related to Proof that every metric space is homeomorphic to a bounded metric space but I can remember that if $d$ is a metric, then $d'(x,y):=\frac{d(x,y)}{1+d(x,y)}$ is also a metric that defines the same topology. I'm stuck on how to prove…
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Is $C[0,1] \setminus P$ (where $P$ is the set of polynomials) connected?

Here's my latest try: Suppose it's not. It means there are two non-empty disjoint open sets $A,B$ such that $A \cup B = C[0,1] \setminus P$. "Take" $f \in A, g \in B$ such that $f(x)
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A problem of checking completeness of some subsets of $\mathbb{R}$

Let $S$ and $W$ be subsets of $\mathbb{R}$, with the usual metric, \begin{align*} S &= \left\{\frac{1}{n} :n\in \mathbb{N}\right\}\cup\{0\} \\ W &= \left\{n+\frac{1}{n}: n\in\mathbb{N}\right\} \end{align*} I have to check for completeness of…
Srijan
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Show that ($\mathbb{R}$, $d$) is a metric space

Question Let $\mathbb{R}$ be the set of real numbers and define $d$ : $\mathbb{R}$ $\times$ $\mathbb{R}$ $\rightarrow$ $\mathbb{R}$ by $d(x, y) = \mid e^{x} - e^{y} \mid $. i) Show that ($\mathbb{R}$, $d$) is a metric space ii) What are the…
Joey
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What is preserved by lipeomorphisms?

A lipeomorphism is a continuous surjective function $f: X \to Y$ between metric spaces such that $$A d_X(x,y)\leq d_Y(f(x),f(y))\leq B d_X(x,y),$$ where $A,B > 0$ are constant. I want to know what kind of metric properties are preserved by this…
EQJ
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Are local quasi-geodesics already quasi-geodesics in hyperbolic spaces?

Recall the following definitions 1) A $(\lambda, \varepsilon)$-quasi-isometric embedding $f$ between metric spaces $X$ and $Y$ is a map $X \to Y$ such that $\frac{1}{\lambda} d_X(x,y) - \varepsilon \leq d_Y(f(x), f(y)) \leq \lambda d_X(x,y) +…
M.U.
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Are there infinitely many non equivalent metric spaces on certain sets (?)

Two metric spaces X and Y are called equivalent if: $d_X (x,x_n) \to 0 \Leftrightarrow d_Y (x,x_n) \to 0 $ with $ n \to \infty $ I wonder whether, if you took a certain set (for example a finite set, the natural numbers, or any other, compact, non…
Imago
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Showing that $d(m,n)=|m^{-1}-n^{-1}|$ is not a complete metric on $\mathbb Z^+$

Let $X$ be a set of all positive integers and define metric $d$ on $X$ by $d(m,n)=|m^{-1} - n^{-1}|$. I'm required to show $(X,d)$ is not a complete space. SOLUTION: Let $\{x_n\}$ be any Cauchy sequence in $X$. Then choose $\epsilon=0.5$; there…
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Good function's

I'm trying to solve the following question: Let $(X,d)$ be a metric space. We call a continuous function $f:X\to \mathbb R$ "good function" if for every continuous function $g:X\to \mathbb R$, the set: $S=\{x\in X;f(x)g(x)=1\}$ be a compact…
hamid kamali
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Completeness of continuous real valued functions with compact support

How can I show that the space of continuous real valued functions on R with compact support in the usual sup norm metric is not complete ? I know that this result can be proved by using the fact that the given space is dense in the space of all…
Ester
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