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  1. A topology is a pair of sets $(X,\tau)$ such that

    • $\tau \subseteq \mathcal{P}(X)$.

    • $X,\emptyset \in \tau$.

    • $\forall S\subseteq\tau.\bigcup S \in \tau$

    • $\forall A,B\in \tau.A \cap B \in \tau$

  2. If $\mathcal{B}$ is a base for the topology $\tau$, then $$\tau=\left\{\bigcup S:S\in\mathcal{P}(\mathcal{B})\right\}$$ ($\tau$ is generated by the union of elements of $\mathcal{B}$)

  3. $\{(a,b)\in\mathcal{P}(\Bbb{R}):a,b\in\Bbb{Q}\}$ is a base for the Euclidean topology on $\Bbb{R}$

  4. $\Bbb{Q}\ne\Bbb{R}$

These four statements cannot be true simultaneously, but every explanation of "second-countable" that I have been offered - taken at as stated - translates directly into these four propositions (aside from the fact that I am using $\Bbb{R}$ as an example, but you get my point).

So either I have discovered an error in the foundations of general topology that somehow eluded mathematicians for ~100 years, or the McTextbook explanation of second-countable spaces uses the words "is," "same," "the," and "base" incorrectly.

I choose to believe the latter. So if "second-countable" doesn't mean what Wikipedia says it means, what does it mean?

Edit

The reason why these statements cannot be simultaneously true is a simple matter of substitution.

The equational definition of the Euclidean topology "the union of open balls" is:

$$E=\langle \{(a,b):a,b\in\Bbb{R}\}\rangle=\left\{\bigcup S:S\in\mathcal{P}(\mathcal{\{(a,b):a,b\in\Bbb{R}\}})\right\}$$

This is acheived by substituting "the set of open intervals" for $\mathcal{B}$ in expression 2.

If $\Bbb{Q}\ne\Bbb{R}$, then

$$\{(a,b):a,b\in\Bbb{R}\}\ne\{(a,b):a,b\in\Bbb{Q}\}$$

Therefore,

$$E\ne\langle\{(a,b):a,b\in\Bbb{R}\}\rangle$$

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    $\begingroup$ Those four statements are all true simulaneously. Why do you think that they cannot be? $\endgroup$ – Brian M. Scott May 18 '20 at 0:19
  • $\begingroup$ @BrianM.Scott Edited $\endgroup$ – R. Burton May 18 '20 at 0:29
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    $\begingroup$ Prove your claim by demonstrating a contradiction from those four statements. $\endgroup$ – William Elliot May 18 '20 at 0:30
  • $\begingroup$ @WilliamElliot See edit. $\endgroup$ – R. Burton May 18 '20 at 0:30
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    $\begingroup$ The implication $X\neq Y\implies \bigcup \{A:A\subset X\}\neq\bigcup \{A:A\subset Y\}$ is false. $\endgroup$ – Reveillark May 18 '20 at 0:32
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either I have discovered an error in the foundations of general topology that somehow eluded mathematicians for ~100 years, or the McTextbook explanation of second-countable spaces uses the words "is," "same," "the," and "base" incorrectly.

Or, your argument is flawed.

Your claim beginning "Therefore" is incorrect: in general, just because $\mathbb{A}\not=\mathbb{B}$ we cannot conclude that $$\{\bigcup S: S\subseteq\mathbb{A}\}\not=\{\bigcup T: T\subseteq\mathbb{B}\}.$$ It's possible that two different collections of sets have the same "set of unions."

The second-countability of $\mathbb{R}$ provides such an example, with $\mathbb{A}$ the set of open intervals with real endpoints and $\mathbb{B}$ the set of open intervals with rational endpoints. For a simpler example, take $\mathbb{A}$ to be the set of all one-element sets of natural numbers and take $\mathbb{B}$ to be the set of all finite sets of natural numbers.


The fact that the set of open intervals with real endpoints and the set of open intervals with rational endpoints generate the same topology is a consequence of the following:

Every open interval with real endpoints is a union of open intervals with rational endpoints.

And this is easy to prove. Suppose $a<b$ are real numbers. Consider the set $$I=\bigcup_{p,q\in\mathbb{Q}\cap (a,b), p<q}(p,q).$$ $I$ is a union of open intervals each of which is contained in $(a,b)$, so $I\subseteq (a,b)$. In the other direction, for each $x\in (a,b)$ we can find rationals $p,q$ with $a<p<x<q<b$; then $x\in (p,q)\subseteq I$. So in fact $I=(a,b)$.

  • Here's another argument which may be easier to parse. Pick sequences of rationals $\alpha=(\alpha_i)_{i\in\mathbb{N}},\beta=(\beta_i)_{i\in\mathbb{N}}$ with $\alpha$ converging to $a$ from above, $\beta$ converging to $b$ from below, and both $\alpha$ and $\beta$ always staying inside $(a,b)$. For example, if $a=\sqrt{2}$ and $b=\pi$ we could take $\beta=(3,3.1,3.14,...)$ and $\alpha=(2, 1.5, 1.42, ...)$. Now show $$(a,b)=\bigcup_{i\in\mathbb{N}}(\alpha_i,\beta_i).$$

From this we can write any union of open intervals with real endpoints as a union of unions of open intervals with rational endpoints - and a union of unions is just a union.


As an interesting aside, note that it's not enough to just observe that $\mathbb{Q}$ is countable and dense in $\mathbb{R}$; or more jargonily, there are separable (= countable dense set) spaces which are not second-countable.

The ones that occur to me offhand are pretty pathological beasties. I'll add a more natural example if I can think of one, but for the moment consider the following construction. Given an arbitrary topological space $(X,\tau)$ consider the space $(Y,\sigma)$ with

  • $Y=X\sqcup\{*\}$, and

  • $\sigma=\{\emptyset\}\cup\{U\cup\{*\}: U\in\tau\}$.

Basically, we take $(X,\tau)$, add a new "special" point, and glue that point to every other point in $X$. The space $(Y,\sigma)$ has a one-element dense subset, namely $\{*\}$, but if $(X,\tau)$ wasn't second-countable then $(Y,\sigma)$ won't be either. Some authors$^*$ call this the "one-point katamarification" of a space.

(By contrast, it's easy to show that second-countability does imply separability: just pick an element from each of the countable basic opens.)

$^*$Specifically, me.

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  • $\begingroup$ Certainly, but the topology generated by the countable base does not contain the open set $(-\pi,\pi)$, the one generated by the uncountable base does. They are not the same topology. $\endgroup$ – R. Burton May 18 '20 at 0:33
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    $\begingroup$ @R.Burton Yes, it does: $$(-\pi,\pi)=(-3,3)\cup(-3.1,3.1)\cup(-3.14,3.14)\cup ...$$ Your claim is incorrect. $\endgroup$ – Noah Schweber May 18 '20 at 0:34
  • $\begingroup$ Can you prove your claim? $\endgroup$ – BigbearZzz May 18 '20 at 0:35
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    $\begingroup$ @NoahSchweber Imagine my relief... thanks. $\endgroup$ – David C. Ullrich May 18 '20 at 0:45
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    $\begingroup$ @R.Burton "It might not be necessary to use the ordering relation to prove that the union of a subset of a countable base contains a member of the standard, uncountable base." Sure; for example, any metrizable space with a countable dense subset is second-countable, and $\mathbb{R}$ (with the usual topology) is indeed metrizable and has a countable dense subset. But not all metrizable spaces come from orderings. $\endgroup$ – Noah Schweber May 18 '20 at 2:22
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Consider the interval $(-\sqrt{2},\sqrt{2})$ in $\mathbb{R}$. There is a sequence $\lbrace x_i\rbrace=\lbrace1,1.4,1.41,1.414,...\rbrace$ of rational numbers which converges to $\sqrt{2}$ from below. The union of the intervals $(-x_i,x_i)$ is the interval ($-\sqrt{2},\sqrt{2})$.

To produce any interval $(x,y)$ for $x,y$ real, $x<y$ we can construct sequences of rationals $x_i,y_i$ where $x_i$ converges to $x$ from above and $y_i$ converges to $y$ from below. The union of the intervals $(x_i,y_i)$ will be the required interval. These intervals form a basis for the usual topology on $\mathbb{R}$.

Note that while the end-points of these intervals are rational, they are intervals in $\mathbb{R}$, not merely in $\mathbb{Q}$.

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