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For a linearly (totally) ordered set $A$, one can define its order topology: that is the smallest topology containing the set B of all the intervals of the form $\{x\mid x < a\}, \{x\mid x > a\}$ or $\{x\mid a < x < b\}$ where $x$, $a$ and $b$ are elements of $A$.

The order topology can be shown to be generated by the set $B$ of all intervals mentioned above (that is any element of the order topology can be written as the union of intervals). But to prove this, isn't AC required?

Otherwise how can we prove that: if we call $U$ the set of all union of indexed families of elements of $B$, then any union of indexed families of elements of $U$ are also elements of $U$? This is required for $U$ to be the order topology.

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The Axiom of Choice does tell us that every set has a topology induced by a well-ordering, but is not necessary here. Indeed, most order topologies that people are interested in do not come from a well-ordering (the metric topology on $\mathbb{R}$, for example, is isomorphic to the topology induced by the dense linear order). – you Mar 5 '12 at 19:56

The axiom of choice is not required. Let $$\mathscr{B}=\{(\leftarrow,a):a\in A\}\cup\{(a,\to):a\in A\}\cup\{(a,b):a,b\in A\}\cup\{A\}\;,$$ and let $$\mathscr{T}=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;.$$

(In case $A$ contains only one point, you do need to include $A$ in $\mathscr{B}$.)

You’re worried about showing that $\mathscr{T}$ is closed under arbitrary unions, so suppose that $\mathscr{V}\subseteq\mathscr{T}$; we want to show that $\bigcup\mathscr{V}\in\mathscr{T}$. For each $V\in\mathscr{V}$ let $$\mathscr{B}_V=\{B\in\mathscr{B}:B\subseteq V\}\;;$$ AC is not required to define this subset of $\mathscr{B}$.

Claim: $V=\bigcup\mathscr{B}_V$.

Proof: Clearly $V\supseteq\bigcup\mathscr{B}_V$. On the other hand, $V\in\mathscr{T}$, so by definition there is some $\mathscr{U}\subseteq\mathscr{B}$ such that $V=\bigcup\mathscr{U}$. Clearly $B\subseteq V$ for each $B\in\mathscr{U}$, so $\mathscr{U}\subseteq\mathscr{B}_V$, and therefore $V=\bigcup\mathscr{U}\subseteq\bigcup\mathscr{B}_V$, and it follows that $V=\bigcup\mathscr{B}_V$. $\dashv$

AC is not used here: we chose just one subset of $\mathscr{B}$, and its existence was guaranteed by the hypothesis that $V\in\mathscr{T}$.

Now let $\mathscr{H}=\bigcup\{\mathscr{B}_V:V\in\mathscr{V}\}\subseteq\mathscr{B}$. Then

$$\begin{align*}\bigcup\mathscr{V}&=\bigcup\left\{\bigcup\mathscr{B}_V:V\in\mathscr{V}\right\}\\ &=\bigcup\bigcup\left\{\mathscr{B}_V:V\in\mathscr{V}\right\}\\ &=\bigcup\mathscr{H}\in\mathscr{T}\;. \end{align*}$$

Again, we’ve not used AC anywhere. AC is also not required in the proof that $\mathscr{T}$ is closed under finite intersections and contains $\varnothing$ and $A$.

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Thank you, that was very clear. – stefan Mar 5 '12 at 18:56

The axiom of choice has little to do with this. Why? The topology is defined from the collection of intervals which is defined by the given order.

We did not have to make any arbitrary choice. We need to verify that the collection of intervals is closed under finite intersections, and from there the topology is simply defined as all the possible unions of intervals.

I think that your question comes from the fact that under the axiom of choice, if we take a family of intervals we can assume the index set is well ordered. This is actually a "misuse" of the axiom of choice, as the axiom of union ensures that any collection (even non-well orderable collections) has a union set.

We start by the collection $\mathcal B=\{(a,b)\mid a<b, a,b\in A\cup\{\pm\infty\}\}$ where $\pm\infty$ denote "the end of the line" from above and below. Indeed the interval $(-\infty,+\infty)=A$.

  • If there are finitely many intervals $(a_i,b_i)$ we can compare all the end points and easily find $(\max a_i,\min b_j)$ is the needed interval. As there are only finitely many intervals we can effectively write a formula which decides which end point give us the intersection.

  • The topology now is defined as the collection of all unions of collections of intervals. Again this poses no problem, consider $\mathcal P(\mathcal B)$, the power set of the collection above. Every element $X$ in this collection is a collection of intervals. By the axiom of union the set $\bigcup X$ exists, and it is a superset of every element of $X$ - so if $X$ is non-empty of course that $\bigcup X$ is non-empty.

  • I claim now that $\tau=\{\bigcup X\mid X\in\mathcal P(\mathcal B)\}$ is a topology. It is closed under unions as $\bigcup_i (\bigcup X_i)$ would be $\bigcup(\bigcup_i X_i)$; and we have finite intersections since $\bigcup X\cap\bigcup Y$ would contain all the elements which are in intervals both in $X$ and in $Y$, that is $\bigcup(X\cap Y)$. Note that $\bigcup\varnothing=\varnothing$ and $\bigcup\mathcal P(\mathcal B)=A$ so $A$ and the empty set are also in $\tau$.

    (Note that I have used the fact that $X_i\in\mathcal P(\mathcal B)$, which is something closed under arbitrary unions. However this is just a simple set theoretical fact, unions of subsets of a certain set are subsets of the same set.)

  • I also claim that this is the order topology. Of course that every interval is open, since $(a,b)\in\mathcal B$ then $(a,b)=\bigcup\{(a,b)\}$. It is left to verify that indeed the open intervals are a basis for this topology, that is every non-empty open set contains an interval. This is trivial, since $\bigcup X$ surely contain every interval which is in $X$.

There was no use of the axiom of choice nowhere along the way. The axiom of choice is needed in topology, but not for the definition of a topology. You can argue that some collections would form a topology only when the axiom of choice is assumed (e.g. all the well-orderable subsets of $\mathbb R$). However in the case of topologies defined naturally by a given structure it is not usually the case where the axiom of choice is needed.

Topology, however, does make heavy use of the axiom of choice. The Tychonoff theorem is equivalent to the axiom of choice; the fact that second countable spaces are Lindelof makes use of some axiom of choice; and there is a plethora of similar examples where it is needed. Those are not uses where one needs to define a topology, but rather study its properties (countability, separation axioms, etc.)

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@stefan: Uhh... what? – Asaf Karagila Mar 5 '12 at 23:22
sorry, I am not familiar with Latex or MathJax... But to go back to your comment, don't you need AC (or similar) to write $$ \bigcup_i (\bigcup X_i) $$ since any an element of $\tau could have several representation? – stefan Mar 5 '12 at 23:28
@stefan: I don't care that there are many different ways to write a certain open set. Do I need the axiom of choice to have $\{0\}=\{1-1,2-2,3-3,4-4,\ldots,\pi-\pi,\ldots\}$? No. I don't need AC for that. I'm not choosing a unique representation, nor I even claim that one exists. – Asaf Karagila Mar 5 '12 at 23:51

The "smallest topology" containing a collection of sets is defined to be the closure of the collection under finite intersection and arbitrary union. Clearly, if you start with intervals in an order topology, then the smallest topology containing them is, since they're already closed under finite intersection, the collection of arbitrary unions. So every set in the order topology must be a union of basic sets.

We haven't used Choice anywhere.

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Well, "the smallest collection ..." need not be a well-defined nothing in the absence of choice. For example it is possible $\mathbb Q$ has several non-isomorphic algebraic closures - despite the fact that it is "the smallest algebraically closed field containing the rationals". Of course that in this case this is not actually a problem, as my answer and Brian's answer show. – Asaf Karagila Mar 5 '12 at 22:20
Yes, but when finding the smallest topology containing a sub-basis, it's just a matter of intersections and unions, nothing complicated and nothing you can't do without Choice. And to my knowledge the resulting topology is unique, again without Choice. – Patrick Mar 6 '12 at 5:10
Patrick, again: in this case it's fine. However arbitrarily saying "the small collection such that..." might not work. One has to show that the definition does not use the axiom of choice - not just saying that it doesn't. – Asaf Karagila Mar 6 '12 at 8:45
@Asaf: I see your point. – Patrick Mar 6 '12 at 13:32
@Asaf: On second thought, although I see that you feel it's necessary to verify by a lengthy proof that choice is not used, it seems obvious to me, even though your proof is nicely written. In my opinion, your first two paragraphs settle the issue. Otherwise, if we're being picky, to use the union axiom you should verify that $\mathcal{B}$ is a set, and that subcollections of $\mathcal{B}$ are sets, but it's clear to me that choice is not needed to do that. A similar issue arises to make sure that intersections are sets. Just my two cents worth. – Patrick Mar 6 '12 at 20:40

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