Results 1 to 3 of 3

Math Help - Definition of topology

  1. #1
    Junior Member HAL9000's Avatar
    Joined
    Aug 2010
    From
    Germany
    Posts
    26
    Thanks
    1

    Definition of topology

    Hello!

    I got somewhat stuck when I read that in the definition of a topology on a set one may drop the first defining property.

    The familiar definition reads as follows:

    A Topology on a set $X$ is a collection \mathcal{T}\subseteq\mathbb{P}(X) of subsets of $X$, called open sets, such that the empty set and the whole of $X$ are in $X$, and \mathcal{T} is closed under finite intersections and arbitrary unions. In symbols:

    (1) $\emptyset,X\in\mathcal{T}$;

    (2) If $O_i\in\mathcal{T},i\in{I}$, where $I$ is a finite index set, then $\bigcap_{i\in{I}}O_i\in\mathcal{T}$;

    (3) If $O_j\in\mathcal{T},j\in{J}$, where $J$ is any (finite or infinite) index set, then $\bigcup_{j\in{J}}O_{j}\in\mathcal{T}$.

    And now it is said that:

    The defining property (1) may be dropped. In fact, $\emptyset \in \mathcal{T}$ follows, if we apply (3) to the empty family, and $X \in \mathcal{T}$ follows, if we apply (2) to the empty family.

    Now I do see that we have to do with an instance of vacuous truth here, because whenever the index set is empty we have nothing to check.

    But on this ground I could have swapped the roles, saying that:
    The defining property (1) may be dropped. In fact, $\emptyset \in \mathcal{T}$ follows, if we apply (2) to the empty family, and $X \in \mathcal{T}$ follows, if we apply (3) to the empty family.
    …couldn't I?

    Am I wrong? Is there possibly more to this than simply addressing to vacuous truth?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by HAL9000 View Post
    Hello!

    I got somewhat stuck when I read that in the definition of a topology on a set one may drop the first defining property.

    The familiar definition reads as follows:

    A Topology on a set $X$ is a collection \mathcal{T}\subseteq\mathbb{P}(X) of subsets of $X$, called open sets, such that the empty set and the whole of $X$ are in $X$, and \mathcal{T} is closed under finite intersections and arbitrary unions. In symbols:

    (1) $\emptyset,X\in\mathcal{T}$;

    (2) If $O_i\in\mathcal{T},i\in{I}$, where $I$ is a finite index set, then $\bigcap_{i\in{I}}O_i\in\mathcal{T}$;

    (3) If $O_j\in\mathcal{T},j\in{J}$, where $J$ is any (finite or infinite) index set, then $\bigcup_{j\in{J}}O_{j}\in\mathcal{T}$.

    And now it is said that:

    The defining property (1) may be dropped. In fact, $\emptyset \in \mathcal{T}$ follows, if we apply (3) to the empty family, and $X \in \mathcal{T}$ follows, if we apply (2) to the empty family.

    Now I do see that we have to do with an instance of vacuous truth here, because whenever the index set is empty we have nothing to check.

    But on this ground I could have swapped the roles, saying that:
    The defining property (1) may be dropped. In fact, $\emptyset \in \mathcal{T}$ follows, if we apply (2) to the empty family, and $X \in \mathcal{T}$ follows, if we apply (3) to the empty family.
    …couldn't I?

    Am I wrong? Is there possibly more to this than simply addressing to vacuous truth?


    This is an excellent example of the very common educational method that I call "Say something

    foggy, explain nothing, see the kids going nuts and go and drink a beer in full satisfaction" .

    In this case the issue is far from being standard in mathematics and different people may have

    different views, though I think that many (most?) people will agree with the book you're working with:

    1) if we take the intersection over an (the, in fact) empty subset of \mathcal{T} , then

    we can think of this as the set of all the elements x\in X that belong to all and each

    of the elements (subsets of X) that appear in that intersection...but there are no

    sets at all there, so we can argue that ANY element in X appears there under the

    logical argument "show me one subset that appears in that intersection and I'll prove you that any

    element of the set is contained there"...

    2) Something simmilar happens with the union of the empty family on \mathcal{T} : if

    there exists an element in that union then some of the subsets there contains it...but there are no

    subsets there so the union is the empty set.

    Yeah, I know...not the most brilliant moment in mathematics foundations (set theory and stuff), but

    if you agree with those conveniences then there's no problem.

    I, for one, would never say such a thing without first trying to explain a little all this madness, and

    even less would I claim that we can drop (1) because so and so, as if it were an obvious fact.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member HAL9000's Avatar
    Joined
    Aug 2010
    From
    Germany
    Posts
    26
    Thanks
    1
    "show me one subset that appears in that intersection and I'll prove you that any

    element of the set is contained there"
    I find the above statement a bit unclear. Let me try to clarify it for myself.

    I think the whole matter here makes use of the ultimate premise that a priori \mathcal{T}\neq\emptyset$, i.e. whenever we define a topology on X, we can be sure that every x\in X is in some open subset.

    Now take the property (2). It says that the proposition " \{x\in X:\forall i \in I (x \in O_i)\} is open" is true whenever " \forall i \in I : O_i is open" is true. But the truth of the former relies solely on the truth of \forall i \in I (x \in O_i), which is the same as i\in I \Longrightarrow x \in O_i. So in case I:=\emptyset it is true iff x \in O_i. But this last proposition is true a priori, because as you said, if one shows up with an open set, then we are always able to assign some x\in X into this open set, and this goes for all elements of X.

    Take the property (3). Again, It says that the proposition " \{x\in X:\exists i \in I (x \in O_i)\} is open" is true whenever " \forall i \in I : O_i is open" is true. The truth of the former depends on the truth of \exists i \in I (x \in O_i)\}, which is the same as i\in I \land x \in O_i. So in case I:=\emptyset it is false. This means that we deal with all those x in X, which are in no open set. But this is a priori not fulfilled by all the elements of X. So the set of all those elements in X, for which "x in X is such that x is in no open set" is fulfilled is empty.

    OK, I hope I didn't mess up the matter any further...
    Last edited by HAL9000; March 5th 2011 at 11:36 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 23rd 2011, 05:43 AM
  2. a topology such that closed sets form a topology
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: June 14th 2011, 04:43 AM
  3. Show quotient topology on [0,1] = usual topology on [0,1]
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 5th 2010, 04:44 PM
  4. topology epsilon-delta definition of continuity
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 23rd 2009, 03:10 AM
  5. Compact Topology Definition
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 12th 2009, 09:22 AM

Search Tags


/mathhelpforum @mathhelpforum