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
0
votes
1 answer

Why is a contraction defined as a function from a metric space M to itself?

Why is a contraction not defined as a function from one metric space $M$ to another one $N$ with the usual property, i.e.: $ \exists k \in (0,1): $ $$ d_{N}(f(x), f(y)) \le k \, d_M(x,y), \quad \forall x,y \in M $$
gen
  • 1,518
0
votes
1 answer

Contraction map mapping an open ball within it

Let $(X,d)$ be a metric space. Let $T: X\to X$ be a contraction map. Then there is a $R>0$ such that $T(B(x, R) )$ is subset of $B(x,R)$. Is it always true? My geometric intuition says the answer is yes. However I cannot proceed…
math is fun
  • 1,142
0
votes
1 answer

suggest study material for p-adic metric and ultra metric etc.

Well,in the book , Elementary Topology Problem Textbook I first found some problems defining p-adic and ultra metric. They have a few problems on it. Now I thought of reading those two types in details but could not find much in any book or on the…
user118494
  • 5,837
0
votes
2 answers

Is {$\phi$} set forms a metric space or not?

Is $\phi$ set forms a metric space or not ? I think, it does not form a metric space, because, we can't specify a metric on $\phi$. But, In many text book, it is not mention that, the set on which, we define metric should be non empty. If I may…
ram
  • 909
0
votes
0 answers

Prove: $\tau(x_{n},x)\rightarrow 0 \iff \tau_{Y}(x_{n},x)\rightarrow 0$

Let $(M,\tau)$ be a metric space and $(Y,\tau_{Y})$ a subspace of a the metric space with an equivalent metric. Prove: for all $\{x_{n}\}\subseteq Y$ and $x\in Y$, $x_{n}\rightarrow x$ according to $\tau\iff x_{n}\rightarrow x$ according to…
gbox
  • 12,867
0
votes
3 answers

Prove $d$ is a metric

Let $d:X\times X\rightarrow \mathbb{R}$ such that: $d(x,y)=0 \iff x=y$ $d(y,x)\leq d(z,y)+d(z,x)$ I have started with symmetry: Let $x=z$ so: $$d(y,x)\leq d(x,y)+d(x,x)$$ $$d(y,x)\leq d(x,y)$$ But I can find how to prove $$d(y,x)\geq d(x,y)$$…
gbox
  • 12,867
0
votes
1 answer

Trouble in solving the converses of two known theorems in metric spaces.

I have two questions regarding metric spaces which I can't solve.These are not homework problems.These came in my mind quite naturally. Here's these : (1) We know that every open (resp. closed) set in the discrete metric space $(X,d)$ is…
0
votes
1 answer

Bijective mapping from a infinite set

How do we define this mapping: Let $A$ be a infinite set and let $B \subset A$. Suppose that $B$ is finite. Show that there exists a bijective mapping $ $ $f: A \rightarrow A\setminus B$ We get a hint: Use denumeration $B = \left\{ b_1, b_2,…
0
votes
0 answers

Little question on nowhere dense subsets of a metric space.

Suppose that $X$ is a metric space with a nowhere dense subset $A$. Say that $x$ is a point in $X \setminus A$. Then it must be the case that there exists an $r>0$ such that $B(x,r) \cap A = \varnothing$? Any hint would be appreciated.
HeMan
  • 3,119
0
votes
1 answer

Consider the metric space $ \mathbb{R}^{n} $ with the usual metric.

Consider the metric space $ \mathbb{R}^{n} $ with the usual metric. Prove that- $ \ \ max_{1 \leq k \leq n}|x_{k}-y_{k}| \leq d(X,Y) \leq \sqrt n \ \ max_{1\leq k \leq n} |x_{k}-y_{k}| $. $$ $$ Let $ x=(x_{1},x_{2},........,x_{n}) \ \ and \ \…
MAS
  • 10,638
0
votes
2 answers

Condition for contraction mapping

For any $x=x_1, \dotsc, x_n$, $y=y_1, \dotsc, y_n$ in $\mathbf E^n$, define $\|x-y\|=\max_{1 \le k \le n}|x_k-y_k|$. Let $f\colon\mathbf E^n \to \mathbf E^n$ be given by $f(x)=y$, where $y_k= \sum_{i=1}^n a_{ki} x_i + b_k$ where $k =1,2, \dotsc,n$.…
0
votes
1 answer

Question related to Metric space

I was reading Metric space and saw that in one book they defined $R$ as a discrete metric space. But I have read early in the same book that R is normed linear space with usual metric $d (x,y)$$=$$||x-y||$. And I read another remark that any…
0
votes
2 answers

Completion intutiton

I am going through theorems that refer to the completion of a metric space. The following theorem states: If $E$ is a metric space there is a complete metric space $\hat E$ and a function $j:E\to \hat E$ that preserves the metric such that $j(E)$…
Pedro Gomes
  • 3,891
0
votes
1 answer

Rectifiability implies continuity

Consider a rectifiable curve $\gamma:[0,a] \rightarrow S$ on a compact and metric space S. I wonder since the path length of $\gamma$ is finite, it implies that $\gamma$ is continuous? Thanks in advance
jaogye
  • 963
0
votes
2 answers

How to show that $x_n = (1+1/n)^n$ is a Cauchy seq?

I was working on a problem where I should demonstrate that not every Cauchy seq. is convergent. So I choosed $(X,d) = (\mathbb{Q}, |\cdot|)$. Thus $x_n = (1+ 1/n)^n$ which we know lies in $\mathbb{Q}$ but it converges to $e \not \in \mathbb{Q}$. But…
Olba12
  • 2,579