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Math Help - what is the topological dimension of an empty set?

  1. #1
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    what is the topological dimension of an empty set?

    The following picture is extracted from Munkres' "Topology" P305.

    I have problem when the space X is the empty set \emptyset. In this case, the only two open covering is \emptyset and \{\emptyset\}. For \emptyset, the only open refinement is \emptyset, which is also an open covering of \emptyset. For \{\emptyset\}, its only open refinements are \emptyset and \{\emptyset\}, both of them are open coverings of \emptyset. Now the problem comes: what is the order of \emptyset and \{\emptyset\}, respectively? or, what is the "some point" in the definition of order?
    Here are some of my understandings:
    (1)The only topology of \emptyset is \{\emptyset\}, which is discrete. According to Ex 1. of this section(P315), it seems that dim \emptyset should be 0?
    (2) \cup\emptyset=\emptyset, so \emptyset is a covering of \emptyset.
    (3)The first-order logic formulation of " \cal B is a refinement of \cal A" is: \forall x(x\in{\cal B}\rightarrow{x} is a subset of some element of \cal A). For \cal B=\emptyset, the left part of \rightarrow would always be false, then this implication is vacuously true, and in turn the whole wff is true, so \emptyset is a refinement of \cal A. Note: this logic is just like what we use to prove that the empty set is contained in any set.
    (4)"open refinement" means that every element in the collection is open. Express it as a wff: \forall x(x\in{\cal B}\rightarrow{x} is open). The same discussion as (3) shows that \emptyset is an open refinement of \cal A. Additionally, \emptyset is an open covering of \emptyset for the same reason.
    (5)Expressed in first-order logic wff, a natural number m+1 is the order of \cal A if the following is true: [\exists x(x\in{X}\wedge n(x)=m+1)]\wedge[\neg\exists x(x\in{X}{\wedge}n(x)>m+1)], where n(x) is a function giving the number of elements of \cal A that contains x, i.e., n(x)=|\{A\in{\cal A}|x\in{A}\}|. When X=\emptyset, \exists x(x\in{X}\wedge ...) is always false, then no natural number can satisfy the definition of order. This is where my problem lies in.
    Another book written by Engelking, "General Topology", says in P385 explicitly that dim \emptyset=-1. I was totally confused.
    Please help me explain (in rigorous logic) what is the topological dimension of \emptyset, if it exists. Thanks!
    Last edited by zzzhhh; August 6th 2009 at 02:36 AM. Reason: re-attach the image
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  2. #2
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    Having never read Munkres (completely), I got lost about half way through what you are writing but here's how I would look at it.

    Generally speaking, the boundary of an n dimensional set has dimension n-1. That is the surface of a three dimensional object is two dimensional and the boundary of a two dimensional object is one dimensional (a curve). The boundary of a curve consists of its two endpoints: and points have dimension 0. The boundary of a finite set of points is the empty set: the dimension of the empty set is defined to be -1 in order to continue that pattern.
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    Thanks for your reply, your post looks similar to the last paragraph of this section, which mentions another inductive definition of dimension identical to the one in my post when the space is compact metrizable. But this is irrelevant to my question. The picture attached contains all related information. I need a rigorous definition and exposition, that is, in language of math -- mathematical logic, of the topological dimension of the empty space, or somewhat more essential, of the order of the empty collection when space X is empty too.
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  4. #4
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    I see a box with a red X in it but there is no "picture attached".
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    I'm very sorry for inconvenience. After shifting to another picture host, Can u see it now?
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    dim  \emptyset=-1

    I think this is a convention. The definition out of Munkres' book, as it stands, cannot be applied to the empty set, because of that existential quantifier:

    <br />
[\exists x(x\in{X}\wedge n(x)=m+1)]\wedge\neg\exists x(x\in{X}{\wedge}n(x)>m+1)<br />
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