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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.

89 votes
4 answers
46k views

Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.

Is the following true? Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$. I ...
gulim's user avatar
  • 933
111 votes
2 answers
41k views

$A$ and $B$ disjoint, $A$ compact, and $B$ closed implies there is positive distance between both sets.

Claim: Let $X$ be a metric space. If $A,B\subset X$ are disjoint, $A$ is compact, and $B$ is closed, then there is $\delta>0$ so that $ |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$. ...
Benji's user avatar
  • 5,910
54 votes
1 answer
19k views

Show that the countable product of metric spaces is metrizable

Given a countable collection of metric spaces $\{(X_n,\rho_n)\}_{n=1}^{\infty}$. Form the Cartesian Product of these sets $X=\displaystyle\prod_{n=1}^{\infty}X_n$, and define $\rho:X\times X\...
Set's user avatar
  • 7,730
51 votes
4 answers
33k views

If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1 + d(x,y)}$ is also a metric

Let $(X,d)$ be a metric space and for $x,y \in X$ define $$d_b(x,y) = \dfrac{d(x,y)}{1 + d(x,y)}$$ a) show that $d_b$ is a metric on $X$ Hint: consider the derivative of $f(t)$ = $\dfrac{t}{1+t}$ b) ...
Mathsstudent147's user avatar
107 votes
4 answers
76k views

Continuous mapping on a compact metric space is uniformly continuous

I am struggling with this question: Prove or give a counterexample: If $f : X \to Y$ is a continuous mapping from a compact metric space $X$, then $f$ is uniformly continuous on $X$. Thanks for ...
the code's user avatar
  • 1,301
55 votes
3 answers
20k views

If every real-valued continuous function is bounded on $X$ (metric space), then $X$ is compact.

Let $X$ be a metric space. Prove that if every continuous function $f: X \rightarrow \mathbb{R}$ is bounded, then $X$ is compact. This has been asked before, but all the answers I have seen prove the ...
combinator's user avatar
  • 1,366
29 votes
3 answers
7k views

Continuity of the function $x\mapsto d(x,A)$ on a metric space

Let $(X,d)$ be a metric space. How to prove that for any closed $A$ a function $d(x,A)$ is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof: $$ |d(x,a) - ...
SBF's user avatar
  • 36.2k
26 votes
5 answers
4k views

Why doesn't $d(x_n,x_{n+1})\rightarrow 0$ as $n\rightarrow\infty$ imply ${x_n}$ is Cauchy?

What is an example of a sequence in $\mathbb R$ with this property that is not Cauchy? I know that Cauchy condition means that for each $\varepsilon>0$ there exists $N$ such that $d(x_p,x_q)<\...
Ashley's user avatar
  • 937
146 votes
16 answers
25k views

What should be the intuition when working with compactness?

I have a question that may be regarded by many as duplicate since there's a similar one at MathOverflow. In $\mathbb{R}^n$ the compact sets are those that are closed and bounded, however the guy who ...
Gold's user avatar
  • 26.9k
114 votes
3 answers
46k views

When is the closure of an open ball equal to the closed ball?

It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set ...
Alex Lapanowski's user avatar
4 votes
2 answers
3k views

Is a separable and metrizable space second countable?

I believe the answer to this question is yes. I was hoping someone would critique my logic. Let X be a separable and metrizable space. Then it has a countable dense subset A. Let B be a basis for X. ...
ThinkConnect's user avatar
92 votes
6 answers
38k views

Not every metric is induced from a norm

I have studied that every normed space $(V, \lVert\cdot \lVert)$ is a metric space with respect to distance function $d(u,v) = \lVert u - v \rVert$, $u,v \in V$. My question is whether every metric ...
Srijan's user avatar
  • 12.6k
31 votes
2 answers
28k views

Is the distance function in a metric space (uniformly) continuous?

Let $(X, d)$ be a metric space. Is the function $x\mapsto d(x, z)$ continuous? Is it uniformly continuous?
user avatar
22 votes
1 answer
19k views

A isometric map in metric space is surjective? [duplicate]

Possible Duplicate: Isometries of $\mathbb{R}^n$ Let $X$ be a compact metric space and $f$ be an isometric map from $X$ to $X$. Prove $f$ is a surjective map.
C. Simon's user avatar
  • 413
150 votes
7 answers
20k views

$\pi$ in arbitrary metric spaces

Whoever finds a norm for which $\pi=42$ is crowned nerd of the day! Can the principle of $\pi$ in euclidean space be generalized to 2-dimensional metric/normed spaces in a reasonable way? For ...

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