Newsgroups: sci.math Subject: Re: closed and bounded => compact? From: rusin@washington.math.niu.edu (Dave Rusin) Date: 27 Jan 1995 16:02:51 GMT In article <3gb19h$keh@hpg30a.csc.cuhk.hk>, eric wrote: >If a set A is compact, it is bounded and closed. I 'd like to know under >what conditions the converse will also hold (i.e. if A is closed and >bounded, it is compact). My conjecture is that A needs to be complete. >Could someone confirm it or disprove it? Thanks for enlightening me. Keep in mind that "bounded" only makes sense in a metric space, and that "closed" implies X is a subset of something else (evey topological space is closed in itself). In the metric space setting, there is a very pretty circle of ideas which are all equivalent: Theorem. Let X be a metric space. Then the following are equivalent: (a) Every open cover of X has a finite subcover (b) Every infinite subset of X has a limit point (c) Every sequence in X has a convergent subsequence (d) X is complete and totally bounded. Here, a space is "totally bounded" if for every e>0 X can be covered with finitely many balls of radius e. Taking e=1 shows this implies boundedness. Taking a discrete metric space with infinitely many points shows the converse implication does not hold. To finagle the concept of "closed" into here, recall that if Z is a complete metric space, a subset X of Z is closed in Z iff it is complete in its own right. Thus for subsets of a _complete_ metric space, X is compact iff it is closed and _totally_ bounded. It is interesting to note that neither completeness nor total boundedness are topological invariants (e.g. R is complete but not the open interval (0,1) to which it is homeomorphic). Nonetheless, the combination of the two hypotheses is a topological invariant! For spaces which are not metric spaces, (a) makes perfect sense, (d) makes no sense, and (b) and (c) have to be defined appropriately. The lines of implication are then less natural, but yes, you can say something. Reference: Singer and Thorpe, "Lecture notes on elementary topology and geometry", Springer UTM. dave ============================================================================== Newsgroups: sci.math Subject: Re: closed and bounded => compact? From: hook@win25.nas.nasa.gov (Edward C. Hook) Date: Fri, 27 Jan 1995 22:04:59 GMT In article , pecora@zoltar.nrl.navy.mil (Louis M. Pecora) writes: |> Please allow some stupid questions: |> |> In article <3gb5fb$ptj@mp.cs.niu.edu>, rusin@washington.math.niu.edu (Dave |> Rusin) wrote: |> |> > In article <3gb19h$keh@hpg30a.csc.cuhk.hk>, |> > eric wrote: |> > |> > >If a set A is compact, it is bounded and closed. I 'd like to know under |> > >what conditions the converse will also hold (i.e. if A is closed and |> > >bounded, it is compact). My conjecture is that A needs to be complete. |> > >Could someone confirm it or disprove it? Thanks for enlightening me. |> > Dave Rusin pretty much took care of saying what can be said, but it's perhaps worthwhile to explicitly point out that an infinite discrete space, metrized by setting the distance between distinct points equal to 1, shows that completeness is *not* enough. |> > Keep in mind that "bounded" only makes sense in a metric space, and that |> > "closed" implies X is a subset of something else (evey topological |> > space is closed in itself). In the metric space setting, |> > there is a very pretty circle of ideas which are all equivalent: |> |> Can we replace bounded by "contained in a compact set" for non-metric |> spaces? Does that make sense? Does it sort of assume what we want to prove? |> We *can* do something like this -- it is certainly true that closed subsets of compact spaces are themselves compact (which follows trivially from the definition) and, conversely, compact subsets of compact *Hausdorff* spaces are closed (which requires some proof, but not much). |> [[snip]] |> |> > |> > Reference: Singer and Thorpe, "Lecture notes on elementary topology and |> > geometry", Springer UTM. |> > |> > dave |> |> Thanks for any enlightenment. |> |> -- |> Louis M. Pecora |> pecora@zoltar.nrl.navy.mil |> |> /* My views and opinions are not those of the U.S. Navy. |> |> Attractors aweigh my lads, attractors aweigh. */ -- Ed Hook | Coppula eam, se non posit Computer Sciences Corporation / NAS | acceptera jocularum. NASA/Ames Research Center | Me? Speak for my employer?...<*snort*> Internet: hook@nas.nasa.gov | ... Get a _clue_ !!! ... ============================================================================== From: elebeau@ens-lyon.fr (Edouard Lebeau) Newsgroups: sci.math Subject: Re: closed and bounded => compact? Date: 28 Jan 1995 15:56:06 GMT >>> Hi all, >>> If a set A is compact, it is bounded and closed. I 'd like to know under >>> what conditions the converse will also hold (i.e. if A is closed and >>> bounded, it is compact). My conjecture is that A needs to be complete. >>> Could someone confirm it or disprove it? Thanks for enlightening me. 1) bounded makes sense only if A is a metric space. 2) a compact metric space is always complete. 3) a bounded closed subset of a normed (real) vector space E is compact if and only if E is finite-dimensional (Riesz' Theorem) Obviously, the result holds for any space you may imbed topologically in a finite-dimensional (real) vector space. It also holds for finite-dimensional vector spaces over a p-adic field. 4) a complete metric space is compact if and only if it is precompact. Unfortunately, I don't remember the precise definition of `precompact' :-) . ============================================================================== From: palais@binah.cc.brandeis.edu Newsgroups: sci.math Subject: Re: closed and bounded => compact? Date: 28 Jan 1995 19:28:26 GMT A metric space is compact iff it is complete and totally bounded. Totally bounded (or "pre-compact") means that for every positive epsilon there exists a finite epsilon-dense subset (or equiv., there exists a finite cover by sets of diameter < epsilon). So, to answer your question, a cosed and bounded subset of a metric space will be compact provided: 1) The space is complete (closed subsets of a complete space are complete). 2) Bounded subsets of the space are totally bounded. The latter happens to be true in R^n, but is not true in general. (For example, in any metric space redefine the metric by letting the distance between two points equal one if in the original metric their distance exceeds one.Then every subset becomes bounded, but unless the whole space was totally bounded to begin with there are non totally bounded subsets.) ==============================================================================