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In the [[topology]] of [[metric space]]s the '''Heine–Borel theorem''', named after [[Eduard Heine]] and [[Émile Borel]], states:
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For a [[subset]] ''S'' of [[Euclidean space]] '''R'''<sup>''n''</sup>, the following two statements are equivalent:
*''S'' is [[closed set|closed]] and [[bounded set|bounded]]
*every [[open cover]] of ''S'' has a finite subcover, that is, ''S'' is [[compact space|compact]].
 
In the context of [[real analysis]], the former property is sometimes used as the defining property of compactness. However, the two definitions cease to be equivalent when we consider subsets of more general [[metric space]]s and in this generality only the latter property is used to define compactness. In fact, the Heine–Borel theorem for arbitrary metric spaces reads:
 
:A subset of a [[metric space]] is compact if and only if it is [[complete space|complete]] and [[totally bounded]].
 
==History and motivation==
 
The history of what today is called the Heine–Borel theorem starts in the 19th century, with the search for solid foundations of real analysis. Central to the theory was the concept of [[uniform continuity]] and the theorem stating that every [[continuous function]] on a closed interval is uniformly continuous. [[Peter Gustav Lejeune Dirichlet]] was the first to prove this and implicitly he used the existence of a finite subcover of a given open cover of a closed interval in his proof. He used this proof in his 1862 lectures, which were published only in 1904. Later [[Eduard Heine]], [[Karl Weierstrass]] and [[Salvatore Pincherle]] used similar techniques. [[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 set|countable]] covers. [[Pierre Cousin]] (1895), [[Lebesgue]] (1898) and [[Arthur Schoenflies|Schoenflies]] (1900) generalized it to arbitrary covers.<ref name="sundstrom_2010">{{cite arXiv |last=Sundström |first=Manya Raman | eprint=1006.4131 |title=A pedagogical history of compactness |class=math.HO |year=2010 |version=v1 | ref=harv}}</ref>
 
== Proof ==
 
'''If a set is compact, then it must be closed.'''
 
Let ''S'' be a subset of '''R'''<sup>''n''</sup>.  Observe first the following: if ''a'' is a [[limit point]] of ''S'', then any finite collection ''C'' of open sets, such that each open set ''U'' ∈ ''C'' is disjoint from some [[Neighbourhood_(mathematics)#Definition|neighborhood]] ''V''<sub>''U''</sub> of ''a'', fails to be a cover of ''S''. Indeed, the intersection of the finite family of sets ''V''<sub>''U''</sub> is a neighborhood ''W'' of ''a'' in '''R'''<sup>''n''</sup>. Since ''a'' is a limit point of ''S'', ''W'' must contain a point ''x'' in ''S''. This ''x'' ∈ ''S'' is not covered by the family ''C'', because every ''U'' in ''C'' is disjoint from ''V''<sub>''U''</sub> and hence disjoint from ''W'', which contains ''x''.  
 
If ''S'' is compact but not closed, then it has an accumulation point ''a'' not in ''S''.  Consider a collection {{nowrap|''C''&thinsp;′}} consisting of an open neighborhood ''N''(''x'') for each ''x'' ∈ ''S'', chosen small enough to not intersect some neighborhood ''V''<sub>''x''</sub> of ''a''. Then {{nowrap|''C''&thinsp;′}} is an open cover of ''S'', but any finite subcollection of {{nowrap|''C''&thinsp;′}} has the form of ''C'' discussed previously, and thus cannot be an open subcover of ''S''.  This contradicts the compactness of ''S''.  Hence, every accumulation point of ''S'' is in ''S'', so ''S'' is closed.
 
The proof above applies with almost no change to showing that any compact subset ''S'' of a [[Hausdorff space|Hausdorff]] topological space ''X'' is closed in ''X''.
 
'''If a set is compact, then it is bounded.'''
 
Consider the open balls centered upon a common point, with any radius. This can cover any set, because all points in the set are some distance away from that point. Any finite subcover of this cover must be bounded, because all balls in the subcover are contained in the largest open ball within that subcover.  Therefore, any set covered by this subcover must also be bounded.
 
'''A closed subset of a compact set is compact.'''
 
Let ''K'' be a closed subset of a compact set ''T'' in '''R'''<sup>''n''</sup> and let ''C''<sub>''K''</sub> be an open cover of ''K''.  Then {{nowrap|''U'' {{=}} '''R'''<sup>''n''</sup> \ ''K''}} is an open set and
:<math> C_T = C_K \cup \{U\} </math>
 
is an open cover of ''T''.  Since ''T'' is compact, then ''C''<sub>''T''</sub> has a finite subcover <math> C_T',</math> that also covers the smaller set ''K''.  Since ''U'' does not contain any point of ''K'', the set ''K'' is already covered by <math> C_K' = C_T' \setminus \{U\}, </math> that is a finite subcollection of the original collection ''C''<sub>''K''</sub>.  It is thus possible to extract from any open cover ''C''<sub>''K''</sub> of ''K'' a finite subcover.
 
'''If a set is closed and bounded, then it is compact.'''
 
If a set ''S'' in '''R'''<sup>''n''</sup> is bounded, then it can be enclosed within an ''n''-box
 
: <math> T_0 = [-a, a]^n</math>
 
where ''a'' > 0.  By the property above, it is enough to show that ''T''<sub>0</sub> is compact.
 
Assume, by way of contradiction, that ''T''<sub>0</sub> is not compact.  Then there exists an infinite open cover ''C'' of ''T''<sub>0</sub> that does not admit any finite subcover.  Through bisection of each of the sides of ''T''<sub>0</sub>, the box ''T''<sub>0</sub> can be broken up into 2<sup>''n''</sup> sub ''n''-boxes, each of which has diameter equal to half the diameter of ''T''<sub>0</sub>.  Then at least one of the 2<sup>''n''</sup> sections of ''T''<sub>0</sub> must require an infinite subcover of ''C'', otherwise ''C'' itself would have a finite subcover, by uniting together the finite covers of the sections.  Call this section ''T''<sub>1</sub>.
 
Likewise, the sides of ''T''<sub>1</sub> can be bisected, yielding 2<sup>''n''</sup> sections of ''T''<sub>1</sub>, at least one of which must require an infinite subcover of ''C''.  Continuing in like manner yields a decreasing sequence of nested ''n''-boxes:
 
:<math> T_0 \supset T_1 \supset T_2 \supset \ldots \supset T_k \supset \ldots </math>
 
where the side length of ''T''<sub>''k''</sub> is {{nowrap|(2&thinsp;''a'')&thinsp;/&thinsp;2<sup>''k''</sup>}}, which tends to 0 as ''k'' tends to infinity.  Then, by [[Cantor's intersection theorem]], the infinite intersection
 
:<math> T_0 \cap T_1 \cap T_2 \cap \ldots \cap T_k \cap \ldots </math>
 
is not empty,  but instead contains some point ''p''&nbsp;∈&nbsp;''T''<sub>0</sub>.  Since ''C'' covers ''T''<sub>0</sub>, then it has some member ''U''&nbsp;∈ ''C'' such that ''p''&nbsp;∈ ''U''.  Since ''U'' is open, there is an ''n''-ball {{nowrap|''B''(''p'') ⊆ ''U''}}.  For large enough ''k'', one has {{nowrap|''T''<sub>''k''</sub> ⊆ ''B''(''p'') ⊆ ''U''}}, but then the infinite number of members of ''C'' needed to cover ''T<sub>k</sub>'' can be replaced by just one: ''U'', a contradiction.
 
Thus, ''T''<sub>0</sub> is compact.  Since ''S'' is closed and a subset of the compact set ''T''<sub>0</sub>, then ''S'' is also compact (see above).
 
== Generalizations ==
The theorem does not hold as stated for general metric spaces.  A metric space (or [[topological vector space]]) is said to have the '''Heine–Borel property''' if every closed and bounded subset is compact.  Many metric spaces fail to have the Heine–Borel property.  For instance, the metric space of [[rational number]]s (or indeed any incomplete metric space) fails to have the Heine–Borel property.  Complete metric spaces may also fail to have the property.  For instance, no infinite-dimensional [[Banach space]] has the Heine–Borel property.  On the other hand, some infinite-dimensional [[Fréchet space]]s do have the Heine–Borel property.  For instance, the space ''<math>C^\infty(K)</math>'' of smooth functions on a compact set ''<math>K\subset\mathbb{R}^n</math>'', considered as a Fréchet space, has the Heine–Borel property, as can be shown by using the [[Arzelà–Ascoli theorem]].  More generally, any [[Nuclear space|nuclear]] Fréchet space has the Heine–Borel property.
 
The Heine–Borel theorem can be generalized to arbitrary metric spaces by strengthening the conditions required for compactness:
:A subset of a metric space is compact if and only if it is [[complete space|complete]] and [[totally bounded]].
This generalisation also applies to [[topological vector space]]s and, more generally, to [[uniform space]]s.
 
Here is a sketch of the "<math>\Rightarrow</math>"-part of the proof, in the context of a general metric space, according to [[Jean Dieudonné]]:
# It is obvious that any compact set ''E'' is totally bounded.
# Let (''x''<sub>''n''</sub>) be an arbitrary [[Cauchy sequence]] in ''E''; let ''F''<sub>''n''</sub> be the closure of the set { ''x''<sub>''k''</sub> : ''k'' ≥ ''n'' } in ''E'' and ''U''<sub>''n''</sub> := ''E'' − ''F''<sub>''n''</sub>. If the intersection of all ''F''<sub>''n''</sub> were empty, (''U''<sub>''n''</sub>) would be an open cover of ''E'', hence there would be a finite subcover (''U''<sub>''n''<sub>''k''</sub></sub>) of ''E'', hence the intersection of the ''F''<sub>''n''<sub>''k''</sub></sub> would be empty; this implies that ''F''<sub>''n''</sub> is empty for all ''n'' larger than any of the ''n''<sub>''k''</sub>, which is a contradiction. Hence, the intersection of all ''F''<sub>''n''</sub> is not empty, and any point in this intersection is an accumulation point of the sequence (''x''<sub>''n''</sub>).
# Any accumulation point of a Cauchy sequence is a limit point (''x''<sub>''n''</sub>); hence any Cauchy sequence in ''E'' converges in ''E'', in other words: ''E'' is complete.
 
A proof of the "<math>\Leftarrow</math>"-part can be sketched as follows:
# If ''E'' were not compact, there would exist a cover (''U''<sub>l</sub>)<sub>l</sub> of ''E'' having no finite subcover of ''E''. Use the total boundedness of ''E'' to define inductively a sequence of balls (''B''<sub>''n''</sub>) in ''E'' with
#* the radius of ''B''<sub>''n''</sub> is 2<sup>−''n''</sup>;
#* there is no finite subcover (''U''<sub>l</sub>∩''B''<sub>''n''</sub>)<sub>l</sub> of ''B''<sub>''n''</sub>;
#* ''B''<sub>''n''+1</sub> ∩ ''B''<sub>''n''</sub> is not empty.
# Let ''x''<sub>''n''</sub> be the center point of ''B''<sub>''n''</sub> and let ''y''<sub>''n''</sub> be any point in ''B''<sub>''n''+1</sub> ∩ ''B''<sub>''n''</sub>; hence we have ''d''(''x''<sub>''n''+1</sub>, ''x''<sub>''n''</sub>) ≤ ''d''(''x''<sub>''n''+1</sub>, ''y''<sub>''n''</sub>) + ''d''(''y''<sub>''n''</sub>, ''x''<sub>''n''</sub>) ≤ 2<sup>−''n''−1</sup> + 2<sup>−''n''</sup> ≤ 2<sup>−''n''+1</sup>. It follows for ''n'' ≤ ''p'' < ''q'': ''d''(''x''<sub>''p''</sub>, ''x''<sub>''q''</sub>) ≤ ''d''(''x''<sub>''p''</sub>, ''x''<sub>''p''+1</sub>) + ... + ''d''(''x''<sub>''q''−1</sub>, ''x''<sub>''q''</sub>) ≤ 2<sup>−''p''+1</sup> + ... + 2<sup>−''q''+2</sup> ≤ 2<sup>−''n''+2</sup>. Therefore, (''x''<sub>''n''</sub>) is a Cauchy sequence in ''E'', converging to some limit point a in ''E'', because ''E'' is complete.
# Let <math>I_0</math> be an index such that <math>\mbox{ }_{U_{I_0}}</math> contains ''a''; since (''x''<sub>''n''</sub>) converges to ''a'' and <math>\mbox{ }_{U_{I_0}}</math> is open, there is a large ''n'' such that the ball ''B''<sub>''n''</sub> is a subset of <math>\mbox{ }_{U_{I_0}}</math> –v in contradiction to the construction of ''B''<sub>''n''</sub>.
 
The proof of the "<math>\Rightarrow</math>" part easily generalises to arbitrary uniform spaces, but the proof of the "<math>\Leftarrow</math>" part (of a similar version with "sequences" replaced with "filters") is more complicated and is equivalent to the [[Boolean prime ideal theorem|ultrafilter principle]],<ref>{{cite book |author = [[Eric Schechter]] |title= Handbook of Analysis and Its Foundations |publisher= Academic Press |year= 1997|isbn= 0-12-622760-8 }}  [http://www.math.vanderbilt.edu/~schectex/ccc/excerpts/equivuf2.gif UF24], p. 506.</ref> a weaker form of the [[Axiom of Choice]]. (Already, in general metric spaces, the "<math>\Leftarrow</math>" direction requires the
[[Axiom of dependent choice]].)
 
== See also ==
* [[Bolzano–Weierstrass theorem]]
 
==Notes==
{{reflist}}
 
== References ==
* {{cite journal | author=P. Dugac | title=Sur la correspondance de Borel et le théorème de Dirichlet–Heine–Weierstrass–Borel–Schoenflies–Lebesgue | journal= Arch. Internat. Hist. Sci. | year=1989 | volume=39 | pages=69–110}}
* {{planetmathref|id=3328|title=proof of Heine-Borel theorem}}
 
== External links ==
* {{cite video |people=Ivan Kenig, Dr. Prof. Hans-Christian Graf v. Botthmer, Dmitrij Tiessen, Andreas Timm, Viktor Wittman |title=The Heine–Borel Theorem |format=avi • mp4 • mov • swf • streamed video |date=2004 |publisher=Leibniz Universität |location=Hannover |url=http://www.math.uni-sb.de/ag/schreyer/oliver/calendar.algebraicsurface.net/calendar.php?mode=youTube&day=07}}
* {{springer|title=Borel-Lebesgue covering theorem|id=p/b017100}}
 
{{DEFAULTSORT:Heine-Borel Theorem}}
[[Category:Theorems in real analysis]]
[[Category:General topology]]
[[Category:Properties of topological spaces]]
[[Category:Compactness theorems]]
[[Category:Articles containing proofs]]

Latest revision as of 01:47, 22 December 2014

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I can go on and on about the benefits of buying, but we are not here for that. Let's get severe about our buying of us, and discover out the perfect streets on the planet to indulge in the one pastime that liberates. And if you love your GUCCI, LOUIS VUITTON, PRADA, and so on.; this Buzzle article aims to acquaint you with the ABC's of buying havens spewed the world over. A procuring oasis in Istanbul, here you can see luxurious retail shops offering each Turkish and international designer labels on either facet of the street. Relaxation those tired legs at exclusive cafés, hotels, and restaurants that dot the road. Abdi Ipekçi Avenue has a popularity of being the most expensive streets in Turkey. The right way to Motivate a Sales Group to Set Parties to Promote Jewelry

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It is not good to purchase used tires because they're typically filled with temporary sealant. Tires which have been used for greater than 6 years have high chance of tread separation while on the street, resulting in fatal accidents than typically result in loss of life. You can not deny the fact that a second-hand tire could have had punctures, suffered wear and tear, ozone cracking, cuts and loss of air pressure on account of weight imbalance. Even improper storage of the tires can contribute to deterioration. You might as properly go and pick up some from a tire graveyard, and find yourself in your personal.

In fact, low cost tires are available far and wide nowadays, with most dealerships providing some actually enticing bargains. Likewise, there are an abundance of internet sites additionally with extremely aggressive provides. Nevertheless, it is vitally important that once you have made the decision to fit discount tires in your motor, that you simply pay particular consideration to components such as the velocity index of the tires , and naturally, also the load ranking. How Are Car Tires Rated? Performance – Efficiency tires are designed to carry to the highway floor under excessive speeds. These tires make use of a softer rubber compound than touring tires or all-terrain tires The softer compound offers better grip and higher performance, however comes at the cost of diminished longevity. Tire Shine

There are new 'stud-much less' tires that apparently are the equivalent of studded tires and seem like getting good critiques. They may catch on out here on the West Coast - the place we won't really use studded tires for the total season as a result of our copious amounts of rain fall. Do not be afraid to ask your native tire retailer for a discount and ensure to price all fitting, set up and a last fitted price to make sure there are not any other extra hidden costs. Your then ready to see if your researched price may be overwhelmed. Summary to Buying Tires When you have an surprising blow-out and are in need of a tire however would rather pay much less or match the tread put on of your different three tires. How Much Do Two Yamaha Dirt Bike Tires Cost? Portray Whitewall Tires to Black

If you want to purchase discount tires on-line at the absolute best value, you must contemplate new and used tire auctions on your tire purchase online. Most of the time, the tires are coming immediately from a manufacturer or retailer surplus. As long as you've got already achieved the research, you can focus in on precisely what you wish to buy. The discount costs on the tires far outweigh any supply you will see anywhere else. If Sears can sell a few of its assets, maybe it may possibly it begin to straighten out its steadiness sheet, but who's going to be considering a majority of these retail property in this market atmosphere until it's a full-blown fire-sale? At fireplace sale costs, are the proceeds going to be sufficient to help Sears balance sheet? Used efficiency tires

These are all easy techniques that can increase lifespan of a automobile, specially in the case of pre-owned vehicles. Every automobile owner should be aware that a automobile is dependable so long as it is sorted. Your car can run for a number of years without any issues, if you happen to incorporate these easy ideas into your long term automobile maintenance technique. The Mustang is a seasoned veteran since its existence started the 1960's by sticking with their nice kinds which have been beloved by vehicle enthusiasts by your entire years. The actual inexpensive pricing plus nice efficiency it gives made the vehicle successful. The brand new fashions of the Kia Mustang will carry on make its point out on Ford's product sales this coming season. Concerning the Creator

The polyester unit makes resin for a wide range of merchandise that embrace beverage bottles, containers for microwave ovens and adhesives. The unit also produces textiles for polyester material that is used to make tires. Tire 'load range', or the maximum weight every tire can safely help, should be considered when choosing the right dimension tire to your application. The load vary and most weight capacity are indicated on the tire facet wall. Tire 'inflation stress' is also an important factor in correct handling in addition to tire life. Most inflation strain is indicated on the tire aspect wall and may always be checked when the tire is chilly earlier than operation. REFERENCE INFORMATION Truck and Bus Sales and Reference Guide An Rationalization of Tire Balance Numbers

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