Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [complete-spaces]

A metric space is complete if, in it, any Cauchy sequence is convergent.

Intuitively, a space is complete if there are no “points missing” from it, as far as limits of sequences are concerned. For instance, the set of rational numbers is not complete, because e.g. $\sqrt{2}$ is “missing” from it, even though one can construct a sequence of rational numbers that converges to it, which is necessarily a Cauchy sequence. It is always possible to “fill all the holes”, leading to the completion of a given space.

When working on a complete space, one can determine that a sequence converges by proving that it is a Cauchy sequence, thereby avoiding the need of actually determining its limit. This is very useful in Analysis.

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Space of bounded continuous functions is complete

I have lecture notes with the claim $(C_b(X), \|\cdot\|_\infty)$, the space of bounded continuous functions with the sup norm is complete. The lecturer then proved two things, (i) that $f(x) = \lim f_n (x)$ is bounded and (ii) that $\lim f_n \in…
Rudy the Reindeer
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Difference between complete and closed set

What is the difference between a complete metric space and a closed set? Can a set be closed but not complete?
ABC
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Two metrics induce the same topology, but one is complete and the other isn't

I'm looking for an example of two metrics that induce the same topology, but so that one metric is complete and the other is not (Since it is known that completeness isn't a topological invariant). Thanks in advance for any hints or ideas.
Pandora
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Showing that if a subset of a complete metric space is closed, it is also complete

Let $(X, d(x,y))$ be a complete metric space. Prove that if $A\subseteq X$ is a closed set, then $A$ is also complete. My attempt: I tried to prove that every Cauchy sequence $(b_n)$ of points of $A$ converges to a point $b\in A$. However could not…
Amadeus
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Totally bounded, complete $\implies$ compact

Show that a totally bounded complete metric space $X$ is compact. I can use the fact that sequentially compact $\Leftrightarrow$ compact. Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$…
Emir
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Every compact metric space is complete

I need to prove that every compact metric space is complete. I think I need to use the following two facts: A set $K$ is compact if and only if every collection $\mathcal{F}$ of closed subsets with finite intersection property has…
QED
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Show that $l^2$ is a Hilbert space

Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$. (a) show that $l^2$ is H Hilbert space. To show that it's a Hilbert space I need to show that the space is…
user235727
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Extension of a Uniformly Continuous Function between Metric Spaces

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces with $(Y,d_Y)$ complete. Let $A\subseteq X$. I need to show that if $f:A\to Y$ is uniformly continuous, then $f$ can be uniquely extended to $\bar{A}$ maintaining the uniform continuity. My attempt at…
Clayton
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Compactness of a metric space

If a metric space $(X,d)$ is compact then for every equivalent metric $\sigma$, $(X,\sigma)$ is complete. This is because, for any Cauchy sequence in $(X,\sigma)$ has a convergent subsequence due to fact $(X,\sigma)$ is a compact metric space, hence…
Sumanta
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Completion of the real numbers

On the real line $\mathbb{R}$ endowed with the Euclidean topology, I may put different metrics, inducing the same topology, but inducing different completions. For example if one considers the standard Euclidean distance you get $\mathbb{R}$ itself,…
Angelo
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Can Hilbert spaces be defined over fields other than $\mathbb R$ and $\mathbb C$?

Let $V$ be a vector space over a field $K$. Suppose further that $K$ has the following structures: $K$ has a subfield $K_{\mathbb R}$ equipped with a field embedding $K_{\mathbb R}\hookrightarrow\mathbb R$, so we can identity elements of…
WillG
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Is locally completeness a topological property?

I know that completeness itself is not a topological property because a complete and a not complete metric space can be homeomorphic, e.g. $\Bbb R$ and $(0,1)$. However, both $\Bbb R$ and $(0,1)$ are locally complete (each point has a neighborhood…
M. Winter
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Another completion of $\overline{\Bbb{Q}}$

$\overline{\Bbb{Q}}$ is the field of algebraic numbers, let $E$ be the set of embeddings $\overline{\Bbb{Q}}\to \Bbb{C}$ and consider the following norm on $\overline{\Bbb{Q}}$ $$\|\alpha\|=\sup_{\sigma\in E} |\sigma(\alpha)|$$ making it a…
reuns
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Dual space of $\mathcal{C}^n [a,b]$.

I just started reading a few days ago about Banach algebras using the Kaniuth's book. In this, it is said that the space $\mathcal{C}^n [a,b]$ of $n$-times continuously differentiable functions is a Banach algebra with the norm $$||f|| :=…
D.R. Cster
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Can this complete metric space be a Banach space?

Let $(S,d)$ be the space of all sequences in $\mathbb{R}$ with the metric $$d(\mathbf{x},\mathbf{y})=\sum_{i=1}^{\infty}\dfrac{1}{2^i}\dfrac{|\xi_i-\eta_i|}{1+|\xi_i-\eta_i|}$$ where $\mathbf{x}=(\xi_i)$ and $\mathbf{y}=(\eta_i)$. This is a complete…
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