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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Facing difficulty in finding a counterexample to prove that the set SL$(n, \Bbb R)$ is not bounded in M$(n, \Bbb R)$ for $n \geq 2$.

Facing difficulty in finding a counterexample to prove that the set SL$(n, \Bbb R)$ is not bounded in M$(n, \Bbb R)$ for $n \geq 2$. Here SL$(n, \Bbb R)$ is the set of all $n \times n$ matrices whose det is $1$.
User8976
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Making $(0,1)$ complete with a metric $d$ which defines the same topology as that of the Euclidean metric

Can we define a metric $d$ on $(0,1)$ such that the topology induced by this metric is same as that of the usual Euclidean metric on this set ?
user217921
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Show that there exists a real valued continuous function $f$ on $X$ such that $f(a) = \alpha$ and $f(b) = \beta $.

Let $a,b$ be two distinct points of a metric $(X,d)$ and $\alpha , \beta$ be any two given real number . Show that there exists a real valued continuous function $f$ on $X$ such that $f(a) = \alpha$ and $f(b) = \beta $.
user202674
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Topologists Sine Curve problem

Topologist's Sine Curve: $A:=\{(x,\sin \frac{\pi}{x}):0
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A metric pace is complete if any disjoint closed sets have positive distance

Let $(X,d)$ be a metric space such that $d(A,B)>0$ for any pair of disjoint closed subsets $A,B\subset X$. Show that $(X,d)$ is complete. Suppose $X$ is not complete. Then there exists a Cauchy Sequence in $X$ which does not converge. Let $x_n$ be…
Learnmore
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Two metrics having the same property

Let $(X,d_1)$ be a metric space. Assume there are $x,y \in X$, such that $$d_1(x,z)\leq d_1(y,z) \quad \forall z \in X\setminus \lbrace x,y\rbrace. \quad (*) $$ I am trying to show that if one defines another metric $d_2$ on X, this property is…
munky
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Metric spaces - prove the closedness of a subset of a continuous function

One question in my book is to prove the continuity of a subset of $C[0,1]$ (continuous real functions on [0,1]) with metric $d(f,g) = sup_{x\in[0,1]} |f(x)-g(x)|$. However, I am not sure on how to continue. The subset in question is $\{f \in C[0,1]…
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Metric (In $\mathbb{R}^2$) $d$ in which $d((3,3),(4,2))>d((3,3),(3,7))$?

I'd like to find a metric in $\mathbb{R}^2$ (Denoted $d$) in which $d((3,3),(4,2))> d((3,3),(3,7))$. Is there such metric? Adding something. I already have a pseudometric which does that (The one of $d((a,b),(c,d))=\min\{|c-a|,|d-b|\}$.
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Equivalent metric making rational complete

I know of no necessary conditions for a metric d to be equivalent to the standard euclidean metric on Reals.Hence, I was facing difficulty in answering the following problem: Does there exist d, equivalent to Standard metric on reals, s.t. it makes…
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Which metric spaces are isometric to $(\mathbb{R}^n, d_E)$?

Which metric spaces are isometric to $(\mathbb{R}^n, d_E)$? Two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$ are called isometric, when an isometry between them exists. An isometry is an isomorphism $\varphi: M_1 \rightarrow M_2$ such that $$\forall…
Martin Thoma
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Non-empty finite point set is closed

Subset of $\Bbb R^2$: My book says that non-empty finite point sets are closed. Why is this? Since it is a finite point set, it necessarily has no limit points within it, since every neighborhood of a limit point has infinite many points in it. So a…
beginner
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Compactly supported function?

I'm doing some self-study and I ran into a situation as follows. Suppose $(X,d)$ is a metric space and $F\subset X$ is compact. For some $\varepsilon>0$ let $V=\{x:d(x,F)<\varepsilon\}$. Is the function…
T. Eskin
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Does topological equivalence of metrics imply strong equivalence?

I know that if $(X,d_1)$ and $(X,d_2)$ are metric spaces and for some positive constants $a,b$ , $ad_1(x,y) \le d_2(x,y) \le b d_2(x,y) $ for every $x,y$ in $X$ , then a subset $A$ of $X$ is $d_1$ open iff $d_2$ open that is $(X,d_1)$ and $(X,d_2)$…
user210961
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Proving $d_1$ is a metric.

If $d$ is a metric on a set X, then $d_1 = \frac{d(x,y)}{1+d(x,y)}$ is also a metric. I have proved the other conditions of being a metric except the triangle inequality. Please help!!
user208334
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Show $f(x)=x\ln x$ is not uniformly continuous

Show $f:(0,\infty)\rightarrow \mathbb{R}, f(x)=x\ln x$ on $(0,\infty)$ is not uniformly continuous. I think that the general way to prove that something is not continuous in a metric space is to let $\epsilon=...$ and show that $\forall\delta>0$,…
Emir
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