Compact subspace in metric space

[facenian]

Is there an easy example of a closed and bounded set in a metric space which is not compact. Accoding to the Heine-Borel theorem such an example cannot be found in $R^n(n\geq 1)$ with the usual topology.

 

[fresh_42]

Is there an easy example of a closed and bounded set in a metric space which is not compact. Accoding to the Heine-Borel theorem such an example cannot be found in $R^n(n\geq 1)$ with the usual topology.

From Wikipedia:

A simple counterexample yields the discrete metric on an infinite set of $X$. The discrete metric is defined by
$d (x, x) = 0$,
$d (x, y) = 1$ for $x \neq y\; , \; x, y \in X$
In this metric, each subset of $X$ is closed and bounded, but only the finite subsets are compact.

Further counterexamples are all infinite-dimensional normalized vector spaces.

 

[lavinia]

Is there an easy example of a closed and bounded set in a metric space which is not compact. Accoding to the Heine-Borel theorem such an example cannot be found in $R^n(n\geq 1)$ with the usual topology.

Take the closed unit disk in the plane with the following metric. If two points lie on the same radius line through the origin, their distance is the Euclidean distance. If they lie on different radii, it is the sum of their Euclidean distances to the origin. This space is bounded since the maximum distance is 2.

Show that it is closed, i.e. every Cauchy sequence converges to a point in the space.

Is it compact?

 

[facenian]

From Wikipedia:

A simple counterexample yields the discrete metric on an infinite set of $X$. The discrete metric is defined by
$d (x, x) = 0$,
$d (x, y) = 1$ for $x \neq y\; , \; x, y \in X$
In this metric, each subset of $X$ is closed and bounded, but only the finite subsets are compact.

Further counterexamples are all infinite-dimensional normalized vector spaces.

I think it will be hard to beat this example on simplicity basis.
In the case of @lavinia 's example I believe the disk should be an open unit disk.
Thank you guys, it's been very helpful

 

[lavinia]

I think it will be hard to beat this example on simplicity basis.
In the case of @lavinia 's example I believe the disk should be an open unit disk.
Thank you guys, it's been very helpful

No. It is the closed unit disk. If it were open then a Cauchy sequence of points that converges to a point of distance 1 to the origin will not converge.

 

[facenian]

No. It is the closed unit disk. If it were open then a Cauchy sequence of points that converges to a point of distance 1 to the origin with not converge.

Of course!, I'm sorry I was thinking what must happen for it not to be compact and responded on the contrary. Thanks again.

 

[lavinia]

Of course!, I'm sorry I was thinking what must happen for it not to be compact and responded on the contrary. Thanks again.

This space is also path connected unlike the example of the infinite discrete space.

Have you found an open cover with no finite subcover?

 

[mathman]

Another simple example Hilbert space. The unit ball is closed and bounded but not compact. A particular Hilbert space would be all infinite series, where the sum of the squares is finite.

 

[lavinia]

Another simple example Hilbert space. The unit ball is closed and bounded but not compact. A particular Hilbert space would be all infinite series, where the sum of the squares is finite.

Can you describe the topology a little more?

 

[mathwonk]

@mathman, I know you meant this, but technically you need to say infinite dimensional Hilbert space. Indeed a normed space is locally compact iff it is finite dimensional, Riesz's theorem. Dieudonne', Foundations of modern analysis, page 109.

 

[member 587159]

Can you describe the topology a little more?

Probably @mathman refers to the space $\mathcal{l}^2 = \{(a_n)_n \in \mathbb{K}^\mathbb{N}\mid \sum |a_n|^2 < \infty\}$, with the norm induced by the innerproduct $\langle (a_n)_n, (b_n)_n\rangle:= \sum a_n b_n$.

I.e., $\Vert (a_n)_n \Vert = \sum |a_n|^2 < \infty$

You take the topology that is derived from the norm.

 

[WWGD]

Yes

Probably @mathman refers to the space $\mathcal{l}^2 = \{(a_n)_n \in \mathbb{K}^\mathbb{N}\mid \sum |a_n|^2 < \infty\}$, with the norm induced by the innerproduct $\langle (a_n)_n, (b_n)_n\rangle:= \sum a_n b_n$.

I.e., $\Vert (a_n)_n \Vert = \sum |a_n|^2 < \infty$

You take the topology that is derived from the norm.

Yes, there are infinite nets without finite subnets, which is an equivalent to compactness: every e-net has a finite subnet. I think also the sequence $e_i:= \delta^i_j$ has no convergent subsequence, since $||e_i-e_j||= \sqrt 2$

 

[member 587159]

Yes

Yes, there are infinite nets without finite subnets, which is an equivalent to compactness: every e-net has a finite subnet. I think also the sequence $e_i:= \delta^i_j$ has no convergent subsequence, since $||e_i-e_j||= \sqrt 2$

In the context of this question, one can also use the well known theorem from functional analysis that in a normed space the unit ball compact is if and only if the space is finite dimensional. Since the space provided by @mathman is infinitedimensional, the unit ball isn't compact, but clearly it is closed (every closed ball is closed as set) and bounded.

 

[mathman]

@mathman, I know you meant this, but technically you need to say infinite dimensional Hilbert space. Indeed a normed space is locally compact iff it is finite dimensional, Riesz's theorem. Dieudonne', Foundations of modern analysis, page 109.

When I first was exposed to Hilbert space, it was only infinite dimensional, i.e. the infinite dimensional extension of Euclidean space.

 

[mathwonk]

well technically a hilbert space is usually defined as any complete inner product space, so finite dimensional spaces qualify.

https://en.wikipedia.org/wiki/Hilbert_space

But it is true some people use the term in a restricted sense, sometimes even to mean only those infinite dimensional ones with a countable dense subset, basically "little l-2". Then Euclidean spaces are essentially just (isomorphic to) spaces of finite sequences of scalars of fixed length n, and "Hilbert space" is (isomorphic to) the space of infinite square - summable sequences. These are also called "separable" (infinite dimensional) hilbert spaces.

So there are three classes of Hilbert spaces, finite dimensional, separable infinite dimensional, and inseparable (necessarily infinite dimensional), and some people only use the term "Hilbert space" for the second class. My favorite book on the topic, which explains it this way, is the one I started out on in college: Spectral Theory, by Edgar Lorch, Oxford Univ Press 1962. In fact he says there on p. 60, as you learned, that "Historically, the term Hilbert space has been reserved for" the case of countably infinite dimensional ones. (dimension here is in the sense of hilbert space dimension, i.e. the cardinality not of a maximal linearly independent set, but of a maximal orthonormal set, since every hilbert space has uncountable linear dimension.)

 

[fresh_42]

When I first was exposed to Hilbert space, it was only infinite dimensional, i.e. the infinite dimensional extension of Euclidean space.

Does that mean in your world there is no zero operator or at least no restriction of the codomain to the image? I think from a categorial point of view to include finite dimensional Hilbert spaces make absolute sense.

 

[mathwonk]

by the way for the OP, in any metric space, the equivalent of compactness is the combined conditions "complete and totally bounded", replacing the weaker Euclidean conditions of "closed and bounded".