I have two things I'm not sure in 100% about them. The first, is $\omega_1$. I have a little "feeling" of it, but if I'll be asked to define it - I don't know where to begin from. Perhaps it is because I don't understand it well enough yet. For example, what are the elements of $\omega_1$? I've been looking for an easy to understand definition of $\omega_1$ - doesn't have to be a formal one - what I'm looking for is an intuition to "feel it better".

Another thing is $\mathbb{R}^{[0,1]}$ - I've seen this in several occasions, but I don't know what it means?


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    $\begingroup$ $\mathbb R^{[0,1]}$ is the set of functions from $[0,1]$ to $\mathbb R$. In general, the set $Y^X$ is the set of functions $X \to Y$. $\endgroup$
    – Benjamin
    Commented Oct 3, 2014 at 11:46
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    $\begingroup$ Do you know what an ordinal is? $\omega_1$ is the set of finite or countable ordinals. $\endgroup$
    – MJD
    Commented Oct 3, 2014 at 12:06
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    $\begingroup$ Can you explain what in the term "The least uncountable ordinal" or "The set of countable ordinals" is unclear? Maybe that can help giving you a better answer. $\endgroup$
    – Asaf Karagila
    Commented Oct 3, 2014 at 22:25
  • $\begingroup$ You should ask about $\mathbb{R}^{[0,1]}$ in a separate question. $\endgroup$ Commented Oct 3, 2014 at 22:31
  • $\begingroup$ @AustinMohr: Or... not ask a duplicate of a thousand questions already. $\endgroup$
    – Asaf Karagila
    Commented Oct 3, 2014 at 22:49

3 Answers 3


While I agree that @Austin Mohr ‘s answer is a good picture of $\omega_1$, it is technically problematic since it is circular. You cannot define an uncountable ordinal in terms of uncountable union since you have not yet defined uncountable or uncountable union. Also, the notation may give the impression that you would generate $\omega_1$ by taking a countable union.

The modern approach as answered by @Arno is correct, but does not highlight what is special about $\omega_1$.

The problem with these two approaches is that they blur the distinction between taking the limit of a sequence and taking the supremum of a set. Wanting to get a "better feel" for what $\omega_1$ might be, you will do well to consider Cantor's method for generating ordinals.

Cantor stated three principles for generating ordinals :

  • 1st Principle - Given an ordinal $\alpha$, there is a least ordinal greater than $\alpha$ called $\alpha + 1$.

The 1st principle is enough to generate what Cantor called the first number class - namely $1, 2, 3, \dots$.

  • 2nd Principle - Given any increasing sequence $\alpha_n$ of ordinals, there is a least ordinal greater than all of the $\alpha_n$, called lim$(\alpha_n)$.

The second principle enables us to generate all of the countable (infinite) ordinals - what Cantor called the second number class : $$ \omega, \omega+1, \dots, \omega+\omega, \dots \omega^2, \dots \omega^{\omega}, \dots, \omega^{\omega^{\omega}}, \dots, \epsilon_0, \dots,\epsilon_{\epsilon_0}, \dots, \alpha, \dots $$

The problem is that one can go on like this forever without generating an uncountable ordinal, and Cantor knew that there were uncountable sets. Therefore something wholly new was needed to get to $\omega_1$, and this meant a third (and final) principle of generation.

  • 3rd Principle - For every set $A$ of ordinals, there is a least ordinal greater than every member of $A$, called sup$(A)$.

This third principle enables us to generate $\omega_1$ as the supremum of all countable ordinals. (Note that $\omega_1$ cannot be generated using the second principle since there are uncountably many countable ordinals and therefore they cannot be arranged in a sequence.)

All ordinals can be generated using Cantor’s three principles. Importantly, the third principle produces natural breaks in the sequence of transfinite numbers giving rise to uncountable cardinalities.

Informally, let’s consider what it takes to count from $0$ to $\omega_1$. Metaphorically, let us imagine that we are climbing the mountain called $\omega_1$. We start off a $0$ and count $\omega$ ordinals to arrive at $\omega$. Counting the next $\omega$ ordinals will get us to $\omega + \omega$. Once we have done this $\omega$ times, we will arrive at $\omega^2$. Continuing in this fashion we pass $\omega^3, \dots, \omega^n, \dots$ and finally arrive at $\omega^\omega$. Here is a pictorial representation of $\omega^{\omega}$.

enter image description here

That’s a lot of counting, but we are still only in the pimply little foothills of mount $\omega_1$ - to be honest, we are still in base camp putting on our climbing boots. Continuing we get to $\omega^{\omega^{\omega}}$, etc., and finally we arrive at the countable ordinal $$\epsilon_0 = \omega^{\omega^{\omega^{.^{.^{.}}}}} $$

a exponential tower of $\omega$ copies of $\omega$. One could say that we have now left the foothills and arrived at the first significant ridge on the mountain, but we still cannot see the top of the mountain. $\epsilon_0$ is significant because it is where the representation of ordinals by Cantor’s Normal Form ends.

Continuing, we count $\epsilon_0 + 1, \epsilon_0 + 2, \dots $ and we pass objects like $\epsilon_1, \epsilon_2, \dots, \epsilon_{\epsilon_0}, \dots$ yet we still cannot see the peak of mount $\omega_1$. I think you can see where this is going. The problem is that we can only ever take countably many steps. In order to arrive at $\omega_1$ we need a wholly new principle, as formalised by Cantor’s third principle of generating ordinals.

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    $\begingroup$ But for the third principle, wouldn't you first have to prove that there's the set of all countable ordinals (or at least a set that doesn't contain a maximal countable ordinal)? Of course the class of countable ordinals exists, but how do you see that it is a set, rather than a proper class? $\endgroup$
    – celtschk
    Commented Jul 2, 2016 at 9:57
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    $\begingroup$ Correction: In my previous comment, replace "doesn't contain a maximal countable" by "isn't bounded by a countable ordinal". $\endgroup$
    – celtschk
    Commented Jul 2, 2016 at 10:06
  • $\begingroup$ @celtschk Yes, that appears to be the case. Perhaps problems such as this explain why Cantor's original formulation is no longer used. However, in understanding the dramatic difference between countable and uncountable, it does the job. Re-reading this answer after so long, it is a bit embarrassing really. I'm obviously no expert on the subject and I think I was just looking for an excuse to include a cool picture. $\endgroup$
    – gamma
    Commented Jul 2, 2016 at 15:49

$\omega_1$ is the first uncountable ordinal, or, equivalently, the set of all countable ordinals. The countable ordinals in turn can be constructed by the following rules:

  1. 0 is a countable ordinal

  2. If $\alpha$ is a countable ordinal, then so is $\alpha + 1$.

  3. If for each $i \in \mathbb{N}$ $\alpha_i$ is a countable ordinal, then $\sup_{i \in \mathbb{N}} \alpha_i$ is a countable ordinal.

So $\omega_1$ is the smallest set closed under these rules.


I carried around the following "picture" of $\omega_1$ when I was a graduate student. If it doesn't help you, ignore it completely.

The finite ordinals are the counting numbers $1$, $2$, $3$, $\dots$.

The first countable ordinal $\omega$ is the supremum of all the finite ordinals. It is a bit like a half-open interval $[1, \omega) = \{1, 2, 3, \dots \}$. You can never attain $\omega$ by repeatedly adding $1$, but you would reach it "in the limit".

Once you've "reached" $\omega$, you can continue adding: $\omega + 1$, $\omega + 2$, $\omega + 3$, $\dots$. The supremum of this sequence of additions takes us to $\omega + \omega$, also known as $\omega \cdot 2$.

If we carry the open interval analogy a little further, we can start to see the beginning of $\omega_1$: $$ [1, \omega) \cup [\omega + 1, \omega \cdot 2) \cup [\omega \cdot 2 + 1, \omega \cdot 3) \cup [\omega\cdot 3 + 1, \omega \cdot 4) \cup \cdots $$

Once you have uncountably-many of these "half-open intervals", you've finally reached $\omega_1$. Put another way, $\omega_1$ is the union of this uncountable family of intervals.


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