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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.

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Intersection of interiors of sets in a partition of $\mathbb{R}^d$

Let $\mathcal{Q}=\{Q_1,...,Q_n\}$ and $\mathcal{P}=\{P_1,...,P_m\}$ be partitions of $\mathbb{R}^d$ with $n<m$. Assume all $Q_k\in\mathcal{Q}$ and $P_i\in\mathcal{P}$ are close and convex sets. ...
Staltus's user avatar
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1 answer
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Which metrics (on vector spaces) can be induced?

Is there a way to classify which metrics defined on vector spaces can be induced by a norm? ie. there exists norm $n: X\to \mathbb{R}$ on the vector space $X$ such that the metric $d(x,y)=n(x-y)$. I ...
HIH's user avatar
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1 answer
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Axiom of Choice in characterizing openness in subspace

Below is the typical characterization of open sets in a subspace $Y$ of a metric space $X$. $E$ is $Y$-open iff there exists an $X$-open $S$ such that $E = S \cap Y$. The forwards direction usually ...
n1lp0tence's user avatar
1 vote
1 answer
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Does this increasing sequence of subsets of a bounded connected metric space $(X,d)$ terminates at some point at $X$?

This might be a silly question; this is where I'm stuck as to whether a metric-bounded set in a connected metric space is uniformity-bounded in the sense of Bourbaki. Let $(X,d)$ be a bounded ...
Noiril's user avatar
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1 vote
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Is every bounded connected metric space totally bounded?

A metric space $(X,d)$ is said to be bounded if it is equal to a ball $B(x,r)$ of it. It is said to be totally bounded if for all $\epsilon>0$ there is a finite covering of $X$ by $\epsilon$-balls. ...
Noiril's user avatar
  • 590
1 vote
1 answer
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Prove equivalent form of Baire's Category Theorem

I'm trying to prove these two statements of Baire's Category Theorem are equivalent: Let X complete metric space. A subset of X is meagre if it can be written as the countable union of nowhere dense ...
Jackson Smith's user avatar
-1 votes
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Weak* topology and probability measures problem [closed]

I'm working on a problem related to the weak* topology of probability measures on a metric space and would appreciate your help. Here's the statement of the problem: Consider $K$ a metric space and $\...
asdhjfhjkla's user avatar
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1 answer
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Are functions from a non metrizable general topological space into $\mathbb{R}$ continous under the ring structure?

That is, would continous functions from a general topological space X be closed under field addition, multiplication and so on? Supposing $X$ is metrizable the proof is pretty doable, and example of ...
Cathartic Encephalopathy's user avatar
0 votes
1 answer
36 views

Reconciling metric and topological neighborhoods

Let $X$ be a metric space. Given a point in $x \in X$, an open neighborhood is more appropriately called an $\epsilon$-ball $N_\epsilon = \{p \in X : d(p, x) < \epsilon\}$, while a topological ...
n1lp0tence's user avatar
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1 answer
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If a pseudonorm function $N$ is continuous in a given topology, does the pseudometric topology formed by $d(x,y)=N(x-y)$ coincide with first topology?

Define $p_n\#= p_n p_{n-1} \cdots p_1$ for $n = 0$ to be $1$, then we have a function: $$ N : \Bbb{Z} \to \Bbb{Z}, \\ N(x) = \left |-1/2 + \sum_{d\ \mid\ p_n\#} (-1)^{\omega d} \sum_{r^2 = 1 \mod d} \...
SeekingAMathGeekGirlfriend's user avatar
1 vote
0 answers
31 views

Partition a metric space into parts with small measure and diameter

Consider $\Omega = [0,1]$ with Borel $\sigma$-algebra and Lebesgure measure $\mu$. It has the property that for any $n\geq 1$, we can partition $\Omega$ into $n$ parts with small measure and diameter. ...
efsdfmo12's user avatar
  • 477
0 votes
1 answer
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Do totally open sets exist?

In the third answer to this question, a justification is given for calling closed sets closed, since they are literally closed under the $\mathbb{N}$-ary operation of taking limits of (convergent) ...
Carlyle's user avatar
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3 answers
79 views

Completeness meaning (complete basis vs complete metric space)

Today my professor started talking about the formalism of QM. We talked about the eigenvectors of a Hermitian operator (over Hilbert space) as a "complete set". He also mentioned briefly ...
R24698's user avatar
  • 389
1 vote
0 answers
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Set distances with cube neighbourhoods

This is a follow up and somewhat of a variant of the question I asked a couple of days ago (see Invariance of set distances with $\varepsilon$-neighbourhoods), and after devoting some research and ...
Maths Matador's user avatar
-1 votes
0 answers
45 views

Is $d$ a metric on the vector space $X$? [closed]

Suppose $X$ is a vector space over a field $\mathbb{K}$ and $d$ is a metric on $X$. From the "usual" definition of a metric, $d:X \times X \longrightarrow \mathbb{R}$. Since $X$ is a vector ...
عبد الرحمن's user avatar

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