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Closure and completeness

  1. Feb 26, 2009 #1
    Hello Everyone,

    first of all my apologies, may be my question is too stupid for a forum on Topology and Geometry, but it's something I was thinking about for a while without getting an answer : What's the actual difference between a Close set and a Complete set? I mean : from an algebraic point of view, we say that a Close set contains the limits of its successions, whereas a complete set contains the limits of its Cauchy successions. But we can show that a succession is convergent if and only if it is a Cauchy succession, so the two concepts should be equivalent, and I don't see the difference between closure and completeness. From a topological point of view, we define a close set as the complement of an open one, but at this point I don't see the connection with the completeness property. I started to taught the actual difference could be in passing from infinite dimension to finite dimension, but I am not able to see clearly why. Can anyone help me?
    Sorry again if the question is not so relevant. Cheers,


    Davide
     
  2. jcsd
  3. Feb 26, 2009 #2
    Hi Davide,

    Closedness is a property a subset of a any topological space can have. It is "extrinsic" because it depends on which space you are considering it to be a subset of, for example the interval (0,1] is closed in (0,2] but not in [0,2] (with the usual topology).

    Completeness, on the other hand, makes sense only for metric spaces (or so-called uniform spaces). It is an intrinsic property.

    Sometimes however, the two notions coincide, for example for subvectorspaces of a Banach-space.
     
  4. Feb 26, 2009 #3
    Yyat,

    You had mentioned that (0, 1] is an open set in the topological space (0, 2] with the topology on (0, 2] inherited from the topology on R.

    Would one way to prove that (0 , 1] is open in (0, 2] be to find a homeomorphism from
    (0, 2] onto (0, 1], then conclude that (0, 1] is open in (0, 2] since (0, 2] is open by definition of a topological space, and homeomorphisms from topological spaces to topological spaces maps open sets to open sets?

    For this particular case, I am thinking the proof could be done by making use of the homeomorphism of scalar multiplication:

    x->bx, with b non-zero and real, and then noting that (0, 1] = 1/2*(0, 2].

    What are your thoughts?
     
  5. Feb 27, 2009 #4
    Hello again guys,

    so, If I have understood well, the question is that, in principle, closure and completeness are concepts different in nature, and it can just happen they do coincide somewhere? (Actually, I didn,t get the example about the interval on R). Many thanks for replying and Cheers


    Davide
     
  6. Feb 27, 2009 #5
    (0,1] is closed in (0,2]!

    This only proves that (0,1] is open in (0,1] itself.

    I would prove that (0,1] is closed in (0,2] as follows:
    (1,3) is open in R.
    By definition of the induced topology (1,2]=(1,3)[tex]\cap[/tex](0,2] is open in (0,2]. (If B is a subset of A, the induced topology on B is given by sets of the form B[tex]\cap[/tex]U, where U is any open set in A).
    Therefore, (0,1] is the complement of an open set (in (0,2]), hence closed by definition.
     
  7. Feb 28, 2009 #6
    Thank you! Sorry about the inaccurate statements: when I went to post, I got the ideas mixed up in my mind.

    The induced topology looks like a real convenient way to prove openness: if you have a topological space X, and C is a subset of B is a subset of X, all you have to do to prove that the subset set C of B is open in B is to show that C can be written as the intersection of B with some open set V in X.

    A homeomorphism from a topological space X to a topological space Y maps subsets of X that are open in X to subsets of Y that are open in Y.

    So, if B is a subset of A, and A and B are homeomorphic, is it still possible that an open set in B is not open in A?
     
  8. Feb 28, 2009 #7
    Yes. For instance take A=[0,2], B=[0,1], then B is open in B but not in A.

    Maybe the following fact will clarify things: If A is an open subset of a topological space X and is given the induced topology, then any subset B of A that is open in A is also open in X. (Try showing this!)
     
  9. Feb 28, 2009 #8
    Yyat wrote: If A is an open subset of a topological space X and is given the induced topology, then any subset B of A that is open in A is also open in X. (Try showing this!)

    I appreciate all of your help, and knowledge.

    I'll give it a try.

    Please let me know if I make a mistake in the proof.

    Proof:

    Let T be a topology on X. Then since A is open in X, then A is contained in T. Let K be the induced topology on A given by defining

    [tex] K = \{A \cap V [/tex]: [tex] V \in T \} [/tex].

    If B is a subset of A, and B is open in A, then B is contained in K, and it follows by the definition of K that

    [tex] B = A \cap V [/tex] for some set V in T.

    Since this V is contained in T, and A is contained in T, then by the axioms of a topological space (which state that topologies are closed under finite intersection), we have

    [tex] A \cap V [/tex] is contained in T.

    That is, [tex] A \cap V [/tex] is an open set in X.

    Since [tex] B = A \cap V [/tex],

    then by the substitution principle, B is open in X, and the proof is complete.
     
  10. Mar 2, 2009 #9
    Your proof is correct. The induced topology K can be siad to be the relative toplolgy on A if A is considered as an open subspace of X.
     
  11. Mar 2, 2009 #10
    Is there a linear space in any union of linear subspaces is a subspace except the trivial subspaces? pls help
     
  12. Mar 2, 2009 #11
    Is there a linear space in which any union of linear subspaces is a subspace except the trivial subspaces? pls help
     
  13. Mar 3, 2009 #12
    I am not sure I am interpreting the question correctly. But as far as I know, given any linear space (vector space) W, the trivial linear subspaces of W are the origin {0} in W, and W itself.

    Given any linear space X, it can be shown that the union of any of the following spaces

    {{0}, X, Y}} where the trivial subspace {0} is the origin in X, and Y is any arbitrarily chosen subspace of X.

    So, there exists no linear space in which any union of linear subspaces is a subspace except the trivial subspaces.

    I have a quick: we know from set theory that the cardinality of the power set P(S) of any set S is strictly greater than the cardinality of S itself.

    With that in mind, is it necessarily true that given any linear space (vector space) Z, the set Sub{Z} of all subspaces of Z is itself a subspace of Z?
     
    Last edited: Mar 4, 2009
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