This is slightly different question. First I need to mention that I am neither a mathematician nor a researcher. As an ordinary student the separation " with and without Axiom of choice " trouble me a lot. But I am really interested to know further.

  1. Every ideal in a ring contained in some maximal ideal.

  2. Every vector space has a basis.

  3. Every linear space can be made into a Normed space.

  4. Product of any collection of compact topological spaces is compact with respect to the product topology.

  5. There exists a subset of $\Bbb{R}$ which is not measurable (Lebesgue Measurable) for an example the famous Vitali Set .

All the above theorem can be proved using Axiom of choice.

The natural question is , what happened if we are not allowed to use AC ?

Without AC, can we get -

  1. An ideal which is not contained in any maximal ideal in some ring.

  2. A linear space without basis.

  3. A linear space in which no norm can be defined.

  4. Product of any collection of compact topological spaces which is not compact with respect to the product topology.

  5. A non measurable subset of $\mathbb{R}$ .

If the answer of above questions are "Yes", then the study will be more complicated . [except for 5), it will be fun to have all subsets of $\mathbb{R} $ are measurable]. I don't know a linear space without any basis will be useful or not ?

Why the proof of such kind of theorem involves " if we assume the Axiom of choice, then... " ?

Why not we take it as granted as an extra axiom and we forget about the situation happening without the AC ?

As we know given any two distinct points we can draw a unique straight line passes through them. But no one ask the question if the Euclid's axiom is not listed as an axiom , can we draw more lines using two distinct points.

My question is very clear, why not we avoid the phrase "if we assume the AC " and take it as granted ?


4 Answers 4


The question "why worry about the axiom of choice" (or similar) has been asked many times (see here, here, here, or here just to get started). The moral of the story is that there are lots of places where we're interested in doing "set theory" where the axiom of choice fails (even if you aren't a set theorist! For instance many algebraic geometers care about set theory internal to a sheaf topos, which frequently doesn't satisfy AC). Lee Mosher's analogy about groups "assuming the commutative property" is a very good one.

As for the answer to your other questions: Yes, many algebraic properties are true if and only if the axiom of choice is. For instance, it's consistent without AC that

  1. There is a ring with a nonzero ideal that is not contained in any maximal ideal (see here for more)
  2. There are vector spaces with no basis (see here for more)
  3. There are vector spaces over $\mathbb{R}$ with no norm (see here for more)
  4. There is a collection of compact spaces whose product is not compact (see here for more, though less than the others)
  5. No nonmeasurable subsets of $\mathbb{R}$ exist (see here for more)

I hope this helps ^_^

  • 4
    $\begingroup$ @S.G Happy to help, but I'm not a "sir" ;) $\endgroup$ Dec 14, 2021 at 16:28
  • 1
    $\begingroup$ I think it is worth mentioning the fifth example is slightly special, namely it depends on the consistence of an inaccessible cardinal, so if someone doesn't believe in its consistence, it's meaningful to try to find a nonmeasurable set without choice. $\endgroup$
    – Lxm
    Dec 16, 2021 at 0:06

Some of the other answers gave you links to the specific questions you've asked. Both the specific propositions, as well as the historic reasons why we put the axiom of choice on a separate status than, say, the axiom of extensionality or power set.

Let me add a particular answer to the question in your title. Why do we keep stating usage of the axiom of choice, if we take it for granted? Well, easy. We don't take it for granted. Yes, the vast majority of mathematicians today, at least those I came across, tend to take the axiom of choice as a given, but it is not taken for granted.

The reason is the same reason why research into the necessity of the axiom of choice, and fragments thereof, is still quite relevant. The axiom of choice is terrible non-constructive. Even if you accept the law of excluded middle, which in itself is not very constructive, the axiom of choice is still on a different level. It really just tells you things exist, without telling you how these things "look like", in the slightest. How does a Hamel basis of $\Bbb R$ over $\Bbb Q$ look like? Does it contain $\pi$? $e$? $\sqrt 2$? The answers to all of these is yes and no, in the sense that if one exists, then many exist, and there's no reason to prefer one to the other.

So even if you take the axiom of choice as a given, it is still something that prevents you from specifying your objects explicitly. If you want to be able to compute your objects, or present them in a very explicit way, relying on the axiom of choice means you can't do that. Specifying that you're using it hints to the reader that it might not be possible; and studying fragments of the axiom of choice lets us better understand what kind of "oracles" we need in order to do that specific computation.

So, yes. We still mention the axiom because it provides information about how explicit we can get with our understanding of our mathematical objects. But you're right. In some contexts, e.g. topology, we can't really get anywhere without using choice, so we just use it without mentioning it at all. My topology professor told us somewhere around week 7, when we reached compactness, that from this point on we have to assume the axiom of choice, and those who don't want to have nothing more to learn; he was right. (I'm guessing he was not aware of the fragments we did use learning about metric spaces, though. But that's besides the point.)

Finally, let me just remark that $\ell^2$, and indeed any infinite dimensional Banach space, will not have a Hamel basis when working in $\sf ZF+DC+BP$, where $\sf BP$ states "every set of reals has the Baire property", but they are still very useful spaces.


Given the broad interest in models of set theory where AC fails, it seems more appropriate to clearly state the hypotheses that you are assuming for the model of set theory in which you wish to work, especially for a theorem which is false in certain models where AC fails.

By comparison, if you had a theorem of group theory that was only true "if we assume the commutative property", you would not dare to leave that hypothesis out, given the broad interest in groups that fail to satisfy the commutative hypothesis.


As HallaSurvivor pointed out, each (most?) of these statements are equivalent to the axiom of choice. However I want to point out that just because $\phi$ is provable from $ZF + \neg AC$ (or consistent with $ZF$) does not imply that $\phi$ is provable in $ZF$. In fact, since choice is independent of $ZF$, $\phi$ is not provable from $ZF$ alone.

So if you are simply "not allowed to use AC", you will not be able to prove nor disprove the existence of any of

  1. An ideal which is not contained in any maximal ideal in some ring

  2. A linear space without basis

  3. A linear space in which no norm can be defined

  4. A non-compact product of compact topological spaces

  5. A non measurable subset of R.

  • 3
    $\begingroup$ I don't understand, how does completeness theorem come into play? And what do you mean by "restricting our language"? You certainly can prove less things without choice in your axiomatic system. $\endgroup$
    – Lxm
    Dec 16, 2021 at 0:00

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