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).

15788 questions
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Metrics whose sum induces a topology different from components

Does there exist a metric $d$ defined as the sum of two metrics $d_1$ and $d_2$ (i.e. $d(x,y) = d_1(x,y) + d_2(x,y)$), but $d$ generates a different topology than both $d_1$ and $d_2$? If so, what are some examples of this?
jh1001
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Is a metric space with unique balls infinite/uncountable?

Let $M$ be a non-empty metric space with the following property: if $B(x,r) = B(y,s)$, then $x=y$ and $r=s$, where $B(x,r)$ is the closed ball with center $x$ and radius $r$. Must $M$ be infinite? In fact, must $M$ be uncountable?
user107952
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Will $x+U$ be open set in metric space $X$ if $U$ is open in $X$?

"If $U$ is an open set in normed linear space $X$, then for any $x\in X$, $x+U$ is open". My question is: What can we say if $X$ is not a normed linear space, will it hold? I tried finding counterexamples but couldn't find it. Any hints?
Lucas
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How to continue the induction demonstration?

Be $\{A_n\}_{n \in \mathbb{N}}$ a sequence of connecteds ones such that $A_n \cap A_{n+1} \neq \emptyset \ \forall n \in {N}.$ Prove that $\cup_{n\in {N}}A_n $ is connected. I tried to define $B_n$ as being the union of $A_1$ to $A_n$. By…
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Are Pringles an example of anti-de Sitter space?

Pringles, the potato chip, have the classic saddle shape of anti-de Sitter space. They seem to be the perfect teaching example of anti-de Sitter space, but I've never heard Pringles used in examples. Is there a reason why Pringles arn't used in…
Zamicol
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How to show C_e is closed and not dense in C.

Let $C_{e}([-1,1],\mathbb{R})$ denote the set of even functions in $C([-1,1],\mathbb{R})$ (a) Show $C_e$ is closed and not dense in $C$. (b) show the even polynomials are dense in $C_e$, but not in $C$. I can't start on it... I can't catch any…
syko
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How can we define a Metric $d$ ( in space $\mathbb{R}$ of reals) so that sequence $A_n = 1/n, n = 1,2,3,...$ converges to a non-zero real number?

How can we define a Metric $d$,in space $\mathbb{R}$ of real numbers,so that sequence $A_n = \frac 1 n$ , $n = 1,2,3,...$, converges to a non-zero real number ?
rjdio
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Prove $\{(x,y,z)\in \mathbb{R^3}:x^2+y^2+z^2=3\}$ is compact set

Prove $\{(x,y,z)\in \mathbb{R^3}:x^2+y^2+z^2=3\}$ is compact set
Peter
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Doubt about sets open in metric subspaces need not be open in the respective metric space.

In the example given the set [0,a),(a,1] is open on set S=[0,1] but closed on R. As there are no examples provided to illustrate and directly proceeded to other theorems, i am trying to decipher. Cant we choose radius 'r' finitely large such that we…
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Distance function differentiable at finite points.

Let $A=[1,2]\cup [3,4]\subset \Bbb{R}$. $x\in \Bbb R$, let $f(x)=\text{inf}\{|x-y|: y\in A\}$. Then the function $f(x)$ is continuous but not differentiable at finitely many point. Thanks!
Unknown
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Nowhere-Dense-Sets Clarified

If A is a nowhere dense set, it means that "A closure" has an empty interior. Canwe also say that "A" too has an empty interior? I believe Yes, we can say so. Because "A closure" is just the union of A and it's limit points. The set of limit points…
Krishan
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Why in Cantor's Intersection Theorem the infinite intersection must contain only one point?

I am going through GF Simmons' Intro to Topology and Modern Analysis. $F_n$ is a decreasing sequence of non-empty closed subsets of the Metric Space. It seems that the condition $d(F_n)$->$0$ given in the hyothesis, ensures that the…
Krishan
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How to show a subset of $C[a,b]$ is open?

I would like to show that $C[0,1] \setminus X = \{f(x) \in C[0,1] : f(x) < x\}$, where our metric is $d_{\infty}(f,g) = \text{sup}\{|f(x)-g(x)| : x \in [0,1]\}$, is open which would imply $X$ is closed but am I unsure as to how to go about showing…
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Continuous function takes discrete value on disjoint closed sets

Let $(X,d)$ be a metric space. $A,B$ be disjoint closed subsets of $X$. Then there must exists a continuous function $f$ on $X$, with $f(A)=0$, $f(B)=1$. My idea is to take combination of distance function to $A, i.e$ $1-d(x,A)$. Then the function…
xyz
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Let $(S, d)$ be a metric subspace of $(M, d),$ and $X\subseteq S$. Then $X$ is open in $S$ iff $X=A\cap S$ for some set $A$ which is open in $M$.

Theorem : Let $(S, d)$ be a metric subspace of $(M, d),$ and let $X$ be a subset of S. Then $X$ is open in $S$ if, and only if, $$ X=A \cap S $$ for some set $A$ which is open in $M$. Proof : Assume $A$ is open in $M$ and let $X=A \cap S$. If $x \in…