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

Set of Limit points sentence

So my book says that if $A\subset X$ is finite then $A'=\emptyset$ does that go backwards aswell? meaning that if $A'=\emptyset$ does that mean that the set is finite? I have an exercise to do that wants me to prove that if $A'=\emptyset$ then prove…
NickSorgas
  • 31
  • 1
0
votes
1 answer

Exercise over the connected component of a point $x$ in a metric space $E$

In a metric space $E$, how to prove that the connected component of a point $x\in E$ is contained in every open and closed set containing $x$.
0
votes
1 answer

is there a metric space where every subset is bounded but that space is not discrete??

we know that in discrete metric space every subset is bounded which means we can find a ball that contains that subset. we only need to take a radius greater that 1. is this the only space that has this property? is there a metric space where every…
Hodiii
  • 33
0
votes
1 answer

If $E_r = \{x \in X \mid f(x) < r \}$ is open for all $r \in \mathbb{Q}$, then show that $E_r$ is open for all $r \in \mathbb{R}$.

If $X$ is a metric space and $f: X \to \mathbb{R}$ a map such that $E_r = \{x \in X \mid f(x) < r \}$ is open for all $r \in \mathbb{Q}$, then show that $E_r$ is open for all $r \in \mathbb{R}$. If $E_r$ is open for all $r \in \mathbb{Q}$ it means…
0
votes
2 answers

Showing a set is closed in the real metric if it is also closed in $d_2$ metric

Above is the exercise. Showing that $S$ is bounded is straightforward by $A$ bounded and the triangle inequality, but I thought that to show S is closed, I would do the usual thing of assuming $w \in \bar S$, then by definition, there is a sequence…
nabu1227
  • 879
0
votes
1 answer

how to create an environment for the discrete metric?

Given the discrete metric. I have to show the different environments (amount of shape $\{x ∈ X: d(x,x_0) < r\}$) for different $x_0$ and $r$. My thoughts: $r$ must equal to $1$ because we have the discrete metric, so the environment must look like…
ggg7
  • 3
0
votes
1 answer

Metric spaces - sets proof

Let $(X,\rho)$ be a metric space and $A\subset X$. Prove that $\partial A = \overline A \cap \overline{X \setminus A} $. I have no idea how to prove that. Please help.
user74200
  • 1,027
0
votes
1 answer

Functions that are metrics on the x-axis but not metrics on R^2 as a whole?

Are there functions that can be defined on $\mathbb{R}^2$, are metrics on the whole x-axis (i.e, the set of all points whose second coordinates are $0$), but not metrics on $\mathbb{R}^2$ as a whole? I am especially wondering about the existence of…
Erin
  • 179
  • 10
0
votes
1 answer

Micheal O'Searcoid, Metric Spaces: Problem 8.5:

There exists a function $f: X\to Y$ between metric spaces $X$ and $Y$ that is not continuous but has the property that, for each closed ball $B$ of $Y$, $f^{-1}(B)$ is closed in $X$. What is an example in support of the above statement? Is this…
Saikat
  • 1,583
0
votes
1 answer

Why do we not define the metric on the sequence space as the larget seperation between the sequence?

Space consists of the set of all (bounded or unbounded) sequences of complex numbers and the metric $d$ defined by $$d(x,y)=\sum_{j=1}^{\infty}\frac{1}{2^j}\frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}.$$ where $x=(\xi_j)$ and $y=(\eta_j).$ I just want…
Unknown
  • 3,073
0
votes
1 answer

A doubt on metric space.

Let $c$ be the set of all convergent sequences $x=(x_n)$, then $c$ is a metric space with metric $$d(x,y)=\text{sup}_{1\leq n<\infty}|x_n-y_n|.$$ I have to check all the conditions of metric space. $0=|x_n-x_n|\geq \text{sup}_{1\leq…
Unknown
  • 3,073
0
votes
1 answer

Is $\mathbb{Z}[\frac{1}{p}]$ a lattice in $\mathbb{R} \times \mathbb{Q}_p$?

So we have that $\overline{\mathbb{Z}[\frac{1}{p}]} = \mathbb{R}$ that $\mathbb{Z}[\frac{1}{p}]$ is dense in $\mathbb{R}$. How can it be that $\mathbb{Z}[\frac{1}{p}]$ under the diagonal map $t \mapsto (t,t)$ is now a lattice in $\mathbb{R} \times…
cactus314
  • 24,438
0
votes
1 answer

If asked to prove that a set $S$ is closed, is it sufficient to prove that the given set is closed in a particular metric space?

To elaborate, if I were asked to prove that a set $S$ is closed, and I know that there exists a particular metric space $(X,d)$ such that $S\subset X$, is it then sufficient to prove that $S$ is closed in $X$, using the particular metric $d$?
0
votes
1 answer

A bounded set is contained in an open ball

Given metric space $ (X, d) $ . Statement : A set $ E \subset X $ is bounded iff $ \exists $ $ x \in E $ and $ r \in R^{+} $ such that $ E \subset B_{r}(x,d) $. In the above statement, the set $ B_{r}(x,d) $ is an open ball. Is $ B_{r}(x,d) = \{y…
0
votes
0 answers

Prove $f(tx)=|t|f(x)\implies(\exists\alpha>0)f(x)\geqslant\alpha\|x\|$

Suppose that $m \in \mathbb{N}$ and $\mathbb{R}^m=\{(x_1, ..., x_m)|x_i\in \mathbb{R}\}$ is the real vector space of $m$-tuples of real numbers. Let $\|\cdot\|:\mathbb{R}^m \to [0, \infty)$ be a norm on $\mathbb{R}^m$. A function $f:\mathbb{R}^m \to…
Alex
  • 1