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
2
votes
1 answer

Showing that the following two metric are equivalent.

Let $(X,d)$ be a compact metric space, and $f \colon (X,d)\to \mathbb{R}$ is a continuous function such that if $x,y \in X$ and $x \neq y$ than $f(x) \neq f(y)$. Let $t \colon X \times X \to \mathbb{R}$ where $t(x,y) = |f(x)-f(y)|$. I want to prove…
Stef M
  • 111
2
votes
0 answers

Intersection of closed balls

How does one find a decreasing sequence of closed balls ( not necessarily concentric) in a complete metric space whose intersection is empty?
2
votes
1 answer

boundedness imply total boundedness

I have a question, we can consider it for metric spaces specifically. I wanted to ask if is there any known property a metric space $X$ may posses such that any closed bounded set will imply it is also totally bounded. Also, is there a special name…
User666x
  • 844
2
votes
2 answers

What is the completion of a metric space $(\mathbb{Q}, |\ \ |)$?

What is the completion of a metric space $(\mathbb{Q}, |\ \ |)$?
david
  • 39
2
votes
1 answer

A problem of compactness and connectedness

Consider the subset $A$ and $B$ of $\mathbb{R}^2$ defined by $A =\{(x, x\sin\frac{1}{x}) :x\in(0,1]\}$ $B = A\cup \{(0,0)\}$ I have to check for compactness and connectedness of $A$ and $B$. Here is my attempt. $A$ is bounded but not closed as 0 is…
Srijan
  • 12,518
  • 10
  • 73
  • 115
2
votes
2 answers

Space $Y\subset C[0,1]$ consisting of all $x \in C[a,b]$ such that $x(a) = x(b)$ is complete.

This is problem from kreyszig functional analysis: I have to show that space $Y\subset C[a,b]$ consisting of all $x \in C[a,b]$ such that $x(a) = x(b)$ is complete. I am struggling with this problem for over an hour. Even I am not able to figure…
Srijan
  • 12,518
  • 10
  • 73
  • 115
2
votes
2 answers

Are the metrics $d_\infty(f,g)=\text{sup}_{x\in[0,1]}|f(x)-g(x)|$ and $d_1(f,g)=\int^1_0|f(x)-g(x)|dx$ equivalent?

The vector space of continuous functions on $[0,1]$ can be given the two metrics $d_\infty(f,g)=\text{sup}_{x\in[0,1]}|f(x)-g(x)|$ and $d_1(f,g)=\int^1_0|f(x)-g(x)|dx$. Are these two metrics equivalent? I know how to show two metrics are equivalent…
user71346
  • 4,171
2
votes
1 answer

Show that $f$ is continuous onto $(X,d_{1})$ iff $f$ is continuous onto $(X,d_{2})$.

Suppose that $d_{1}$ and $d_{2}$, are equivalents metrics onto a non-empty set $X$, and consider a function $f:X\to Y$, where $(Y,d)$ is a arbitrary metric space. Show that $f$ is continuous onto $(X,d_{1})$ iff $f$ is continuous in $(X,d_{2})$. My…
2
votes
1 answer

Length structure topology

Let $(X,T)$ be a Hausdorff topological space. Suppose $\emptyset\neq A\subset C(X)$, where $C(X)$ is the set of curves in $X$, that is, the set of continuous maps $\alpha:I\to X$, where $I\subset\mathbb{R}$ is an interval (more specifically a…
2
votes
2 answers

Let $(X,p) $ , $ p(x,y)=\sup\limits_k\dfrac{|x_k-y_k|}{1+|x_k-y_k|} $. Show that $(X,p)$ is complete.

Let $(X,p) $ be the metric space of the real bounded sequences with the metric $$ p(x,y)=\sup\limits_k\dfrac{|x_k-y_k|}{1+|x_k-y_k|} $$ Show that $(X,p)$ is complete. I know the metric space of the real bounded sequence with the metric…
aris
  • 21
2
votes
1 answer

Show that $f$ transform complete subspaces of $M$ in complete subspaces of $N$.

Let $f:M\to N$ continuous, such that exist $c>0$ with $d(f(x),f(y))\geq cd(x,y)$ for all $x,y\in M$. Show that $f$ transform complete subspaces of $M$ in complete subspaces of $N$. I know that an application uniformly continuous $f:M\to N$,…
2
votes
1 answer

Metric not induced by any norm

I'm stuck with two problems, both of them involving metric not induced by any norm. 1) $d_p:\mathbb{R}^2 \times \mathbb{R}^2 \longrightarrow \mathbb{R}$ $d_p(x,y) = \left\{ \begin{array}{ll}||x||_2+||y||_2 & \mbox{if } x \neq y \\ 0 & \mbox{if }…
fferrin
  • 121
2
votes
2 answers

Doubt in the definition of distance $d(x,A)$ between a point $x$ in and a subset $A$ of a metric space

If $ (X,d) $ is a metric space and $A$ is a subset of $X$, then, for $x$ belonging to $X$, we define $$d(x,A)= \inf \{d(x,a) : a \in A\}.$$ My question is, Why do we use infimum in the definition, and why not minimum, average, supremum, etc.?
2
votes
4 answers

Show that the rationals are an incomplete metric space without reference to reals

I know that you can create rational sequences that converge to irrationals, but is there a simple way to do this without explicit assumption of the existence of the reals? I'm thinking of something along the lines of 1) Show that there exists a…
ignoramus
  • 475
2
votes
2 answers

Normed vector space, show that $d(a,c)=d(a,b)+d(b,c)$.

In a normed vector space E, if $c-a=t(b-a)$, with $t\geq 1$, then $d(a,c)=d(a,b)+d(b,c)$. I see this problem like, a vector with extremes , a and c, then $c-a=t(b-a)$ is a separation of the segment $\widehat{ac}$. But I don't see the relation…