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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Does ordered set hold between two different metrics

I've come to a problem I'm not quite sure how to tackle, but I realise it's quite simple in its core. Let's assume $(M,d_2)$, where M is a set of vertices in $R^2$ space, and d being Euclidean metric. Now let's assume another metric space $(M,d_1)$,…
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Given a measure $\mu$ on some $\sigma$-algebra $A$, prove that $d(x,y)=\mu((x-y)\cup(y-x))$ is a metric.

Given a measure $\mu$ on some $\sigma$-algebra $A$, prove that $d:A\times A\to[0,\infty)$ defined as $d(x,y)=\mu((x-y)\cup(y-x))$ is a metric. I started by noticing that $d(x,y)=\mu(x-y)+\mu(y-x)$ because of disjoint set property of measures.…
Garmekain
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Show that for $x\in X'$ there exists $\lambda_x\in\Lambda$ such that $B(x,\delta)\subset O_{\lambda_x}$ in a metric space

Question: Let $\{O_{\lambda}:\lambda\in\Lambda\}$ be an open cover of the metric space $(X,d).$ If $X',$ the set of all accumulation points of $X$ is compact then show that $\exists~\delta>0$ such that for $x\in X'$ there exists…
Jave
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Proving continuity of a mapping of a space onto the unit sphere?

It seems obvious that in an arbitrary normed space $(X, \|\cdot\|)$ a mapping $T$ defined as $$ T(x) = \begin{cases} \frac{x}{\|x\|} & t \neq 0 \\\\ 0 & t = 0 \end{cases} $$ is continuous everywhere but 0. However, I'm having…
bosmacs
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Closure and Interior of subsets on $\mathbb{R}^2$

$A = \{ (x,0) | x \in \mathbb{Q} \} $ Now I can tell that the interior is empty and I think that the closure will be the set $A$ but with $\mathbb{R}$ instead of $\mathbb{Q}$ as $\mathbb{Q}$ is dense in $\mathbb{R}$ but I don't know how to write…
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Metric with stronger constraint on the triangle inequality

I have two functions {d1,d2}, where d1 satisfies all metric properties, but one: triangle inequality. However, the following holds: d1(a,b) ≤ d2(a,c) + d1(b,c) Can I say d1 is a kind of metric? Is there a formal definition for it? I know a super…
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Compactness in Hilbert space $\ell^2$

If we have a compact set $A$ contained in $\ell^2$ (Hilbert Space), how can we show that $A$ is closed and bounded?
mimi
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Question about this proof that sequentially compact metric space implies totally bounded

My text book gives a proof that a seq. compact metric space is totally bounded, and I was wondering if someone could answer my question about it? To start, a metric space is seq. compact if every sequence in some metric space $X$ has a convergent…
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Normed vector space $l^{2}$

We have a normed vector space $l^{2}$ of Hilbert. How can we show that 1) if A contained in $l^{2}$ is compact, then A is closed and bounded. 2) in $l^{2}$, there exists a closed and bounded subset that is not compact
mimi
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Set of continuous function defined on some segment $[0,a]$: completeness

Let $S$ be the set of continuous function $f$ defined on some segment $[0,a_f], a_f \ge 0$, and such that $f(0)=0$. For $f$ and $g$ in $S$, let $$ c_{fg}=\max\{z \in [0,a_f \land a_g] : f(x)=g(x) \text{ for all } x \in [0,z] \}, $$ then, define the…
user14108
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Totally bounded subset

We have a metric space $\boldsymbol{(X,d)}$ and $\boldsymbol Y$ is included in $\boldsymbol{X}$, a subset. How can we prove that $\boldsymbol Y$ is totally bounded if and only if the closure of Y is totally bounded? Any help would be greatly…
mimi
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$Y \subseteq(X,d)$ , Does totally bounded set imply that the set is bounded?

$Y \subseteq(X,d)$ If $Y$ is totally bounded then for each $\epsilon$ there is $\{ y_1,y_2,...,y_n\}$ such that $Y \subset \cup_i^n B(y_i,\epsilon)$. Now let's say that I want to bound the set Y. I can choose $\epsilon = M $ and there exist un point…
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How can I show that triangle inequality holds for this metric?

To start I want to say, this is for homework. Please don't give me the answer, I'm just looking for a little help. Let $\mathbb{P}_1[-1,1] = \{f:[-1,1] \to \mathbb{R}: f(t)=a+bt\}$. I need to show that the function defined by: $d(f,g) = \text{sup}…
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How do I show that a function is not continuous?

The definition my textbook gives for continuity is of a function $f: X \to Y$ is: $f$ is continuous $\iff \forall x,x' \in X, \forall \epsilon > 0, \exists \delta > 0: d_X(x,x')<\delta \implies d_Y(f(x),f(x')) < \epsilon$. To show that a function…
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Distance between continious maps

I´m thinking about why the distance for the set $\{f : [a,b] \to \mathbb{R} : \mbox{continious and bounded}\}$ is define as $$d_{\infty}(f,g) := sup_{x \in [a,b]}|f(x)-g(x)|$$ meanwhile for $\{f : [a,b] \to \mathbb{R} : \mbox{continious}\}$ is given…
LH8
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