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Let for example the set be $X=$ {a,b,c}, where a,b,c are of undeterminated nature, the following topology:

$T=$ {{a},{a,b}}, is a topology in the definition proposed by F. S. George in the book Introduction to topology and modern analysis, the definition is:

A topology $T$ of a set $X$ is a collection of subsets of $X$ satisfying the following:

1) The union of a subclass os sets in T is a set in T 2) The finite intersection of sets in T is a set in T

and in the following pages he propose that the empty set and the set X are always in any topology, and the justificative is because one is the union of an empty subclass and the other is a intersection of an empty subclass of $T$, it is not right is it? How do you guarantee that the empty set is always a subclass of a topology? Is because of this that usualy we require that a topological space must have the empty set and the full set instead of only these 2 previous statements, or not?

Andrés E. Caicedo
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Yassin Rany
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  • If you define $\cap X$ by $y\in \cap X\iff \forall x\in X;(y\in X)$ then $\cap \emptyset$ is the set of everything. That is, $\forall y;(y\in \cap \emptyset). $ So if $B$ is a family of subsets of $X$ and $\cup B\ne X,$ you won't get $X $ as a member of the topology unless $X$ is the set of everything..... Consider the case $\cup B=X^\subsetneqq X$.If you have presented the contents of the book correctly, then it says that the topology generated by $B$ should be a topology on $X$ and also on $X^,$ which is absurd. – DanielWainfleet Oct 28 '16 at 20:04

2 Answers2

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No, $T=\big\{\{a\},\{a,b\}\big\}$ does not satisfy that definition. $\varnothing\subseteq T$, but $\bigcup\varnothing=\varnothing\notin T$. This is straightforward: $x\in\bigcup\varnothing$ if and only if there is a $U\in\varnothing$ such that $x\in U$, and since there is no $U\in\varnothing$ at all, there is no $x\in\bigcup\varnothing$.

Moreover, since we’re working in $\wp(X)$, $\bigcap\varnothing$ is the least upper bound of $\varnothing$ in $\wp(X)$, which is $X$. This is a little trickier and is partly a matter of convention. It’s easy to see that if $X$ is any set, and $\mathscr{A}$ is any non-empty collection of subsets of $X$, then $\bigcap\mathscr{A}$ is the largest subset of $X$ (with respect to $\subseteq$) that is contained in every member of $\mathscr{A}$. If we’re working with subsets of a fixed set $X$, as we are when we talk about topologies on $X$, we extend this observation to the case $\mathscr{A}=\varnothing$ and say that $\bigcap\varnothing$ is the largest subset of $X$ that is contained in every member of $\varnothing$. But $\varnothing$ has no members, so this is simply the largest subset of $X$, i.e., $X$ itself.

Brian M. Scott
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Usually we do indeed explicitly require that the emptyset and the whole space be in the topology. However, his argument is right with one caveat.

The union of a family $\mathcal{F}$ is defined as $\bigcup \mathcal{F}=\{x: \exists A\in\mathcal{F}(x\in A)\}$. If $\mathcal{F}$ is empty, this is just $\emptyset$; so the union of the empty family is indeed the emptyset. (This is similar to how the product over an empty set of reals is $1$, and the sum over an empty set of reals is $0$.)

Similarly, the intersection of a family $\mathcal{F}$ is defined as $\bigcap\mathcal{F}=\{x: \forall A\in\mathcal{F}(x\in A)\}$. If $\mathcal{F}$ has no members, then every $x$ satisfies this; so $\bigcap \mathcal{F}$ is indeed the whole space.

What about that caveat I mentioned at the beginning? Well, it has to do with the intersection of the empty family. Strictly speaking, in order to make this work we need to restrict our attention to sets of points in the space $X$: so that $\bigcap\mathcal{F}=\{x\in X: \forall A\in\mathcal{F}(x\in A)\}$. Why? Well, otherwise the intersection of the empty family contains everything - all the points in $X$, all real numbers, all polynomials in seventeen variables, etc. In the language of set theory, the intersection of the empty family is not a set, but a proper class - the class of all sets!

So strictly speaking the line about intersections isn't quite right, in the context of set theory. However, usually all our quantifiers are implicitly bounded - e.g. here by $$\mbox{$\{x:$}$$ we really mean $$\mbox{$\{x\in X:$}$$ - so this isn't a huge deal.

Bottom line:

  • The union of the empty family is the emptyset, full stop.

  • The intersection of the empty family is the whole space, if we use the space as our "universal set", and otherwise isn't even a set; so this is a bit messy, set-theoretically.

  • It's probably a good idea to explicitly include the whole space and the emptyset in the definition of a topology, just for clarity.

Noah Schweber
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  • Just by curiousity, i had never seen sum and product in terms of sets, even with real elements, there is a name for such manipulation? where i can find it?? thank you very much! – Yassin Rany Oct 28 '16 at 20:09
  • @YassinRany See e.g. http://math.stackexchange.com/questions/110546/what-is-the-product-of-the-empty-set. I don't think there's a special name; the point is it does not matter what order you take the sum/product of a finite set in, so we can define the sum/product of an infinite set as the limit of the sum/product of the finite subsets, if that exists. (E.g. $\sum_{i\in A}x_i=S$ iff $\forall\epsilon\exists T\subseteq S$ such that $T$ is finite and for all finite $T'$ with $T\subseteq T'\subseteq S$, we have $\vert\sum_{i\in T'}x_i-S\vert<\epsilon$. So e.g. $\sum_{i\in\mathbb{N}}2^{-i}=1$.) – Noah Schweber Oct 28 '16 at 20:19
  • So in general, to do this we're relying on two properties: (1) the operation involved (sum/product) is associative and commutative, so its value on a finite set of inputs is well-defined, and (2) we have some way of defining "limits" (e.g. a metric) on the numbers we're using as inputs/outputs (here, real numbers). – Noah Schweber Oct 28 '16 at 20:21