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Does Heine Borel theorem hold in a general metric space justify?

Does Heine Borel theorem hold in a general metric space justify?

Many metric spaces fail to have the Heine–Borel property, such as the metric space of rational numbers (or indeed any incomplete metric space). Complete metric spaces may also fail to have the property; for instance, no infinite-dimensional Banach spaces have the Heine–Borel property (as metric spaces).

What is the statement of Heine Borel Theorem?

Heine-Borel Theorem (modern): If a set S of real numbers is closed and bounded, then the set S is compact. That is, if a set S of real numbers is closed and bounded, then every open cover of the set S has a finite subcover.

Is every closed and bounded set compact?

A set K ⊆ R is compact if and only if it is closed and bounded. To show that K is bounded, suppose that K is unbounded. Then for every n ∈ N there is xn ∈ K such that |xn| > n. Since K is compact, the sequence (xn) has a convergent, hence bounded, subsequence (xnj ).

What is compactness for metric space?

A metric space X is compact if every open cover of X has a finite subcover. 2. A metric space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X. [0,1] is sequentially compact (applying Heine-Borel).

Are 1 is complete metric space?

In a space with the discrete metric, the only Cauchy sequences are those which are constant from some point on. Hence any discrete metric space is complete. For example, the sequence (xn) defined by x0 = 1, xn+1 = 1 + 1/xn is Cauchy, but does not converge in Q. (In R it converges to an irrational number.)

Is every compact metric space complete?

Every compact metric space is complete, though complete spaces need not be compact. In fact, a metric space is compact if and only if it is complete and totally bounded.

Is every closed set bounded?

The integers as a subset of R are closed but not bounded. Also note that there are bounded sets which are not closed, for examples Q∩[0,1]. In Rn every non-compact closed set is unbounded.

What is a cover in topology?

In mathematics, particularly topology, a cover of a set is a collection of sets whose union includes as a subset.

How do you show metric space?

1. Show that the real line is a metric space. Solution: For any x, y ∈ X = R, the function d(x, y) = |x − y| defines a metric on X = R. It can be easily verified that the absolute value function satisfies the axioms of a metric.

Is the Heine-Borel theorem true for general metric spaces?

The Heine–Borel theorem does not hold as stated for general metric and topological vector spaces, and this gives rise to the necessity to consider special classes of spaces where this proposition is true. They are called the spaces with the Heine–Borel property.

Can a Banach space not have the Heine Borel property?

Complete metric spaces may also fail to have the property, for instance, no infinite-dimensional Banach spaces have the Heine–Borel property (as metric spaces). Even more trivially, if the real line is not endowed with the usual metric, it may fail to have the Heine–Borel property.

Which is stronger the fan theorem or the Heine Borel theorem?

Using the open-cover definition of ‘compact’, Theorems 0.3 and 0.4 are equivalent to the fan theorem (and so hold in intuitionism ), but Theorems 0.5 and 0.6 are stronger and Brouwer could not prove them, leading him to define ‘compact’ (for a metric space) to mean complete and totally bounded.

Who was the first person to prove the Heine-Borel theorem?

Émile Borel in 1895 was the first to state and prove a form of what is now called the Heine–Borel theorem. His formulation was restricted to countable covers. Pierre Cousin (1895), Lebesgue (1898) and Schoenflies (1900) generalized it to arbitrary covers.