I'm pursuing Set Theory by Enderton and am having trouble understanding the following idea.

Early in the book, the author constructs an "informal view" of sets, which he says he will refine further (fine by me). In it he asks us to imagine a bucket-like diagram, with each hierarchical layer representing a set. So the bottom-most layer is the set $V_0$ (which is the set of all "atoms" from which other sets will be constructed), the next one is $V_1$, and so on. He then defines a layer as

$$ V_{\alpha+1} = V_{\alpha} \cup \mathcal P (V_{\alpha}) $$

which means that $V_0 \subseteq V_1 \subseteq V_2 \ldots $

Now comes the knockout punch: "But even this infinite hierarchy does not include enough sets. For example, $\phi \in V_1$, $\{\phi\} \in V_2$, $\{\{\phi\}\} \in V_3$, etc., but we do not yet have the infinite set $\{\phi, \{\phi\}, \{\{\phi\}\}, \ldots\}$. To remedy this lack, we take the infinite union $V_{\omega} = V_0 \cup V_1 \cup V_2 \ldots$ and then let $V_{\omega+1} = V_{\omega} \cup \mathcal P(V_{\omega})$."

My objection is: whatever the set $V_0$ is, $\mathcal P(V_1)$ will have $\phi$ as its member, and then $\mathcal P(V_2)$ will have $\{\phi\}$, $\mathcal P(V_3)$ will have $\{\phi, \{\phi\}\}$, and so on. Eventually there has to be a set which contains $\{\phi, \{\phi\}, \{\{\phi\}\}, \ldots\}$. What, then, does the infinite union introduced above achieve?

Also, the idea is later used to demonstrate that there is no set of all sets. Can someone show how this follows (I can add more details if needed)?


1 Answer 1


Regarding your first question: it is not true that "eventually, there has to be a set which contains $\{\phi, \{\phi\},\{\{\phi\}\},\ldots\} \in V_\alpha$." The $V_\alpha$ constructed so far are all constructed using a finite number ($\alpha$) of steps. So even though you have constructed $V_\alpha$ for arbitrarily large $\alpha$, each one is a finite number of steps away from $V_0$.

More carefully: Pick any $\alpha$. The set $\{\phi, \{\phi\},\{\{\phi\}\},\ldots\}$ contains $\overset{\alpha+1}{\overbrace{ \{\cdots\{}} \phi \overset{\alpha+1}{\overbrace{ \}\cdots\} }}$, which is not an element of $V_\alpha$. So $\{\phi, \{\phi\},\{\{\phi\}\},\ldots\} \not\subseteq V_\alpha$.

Regarding the non-existence of the set of all sets, I suppose the same kind of argument demonstrates that, at any level of this construction, there is always a next level. So if you think you have constructed the set of all sets, you can always union it with its power set to construct a set which isn't an element (hence you didn't have the set of all sets). The most common proof of the non-existence of the set of all sets, though, is Russell's Paradox, which demonstrates that its existence yields a paradox.

  • $\begingroup$ Thank for the explanation of part 2 but for the initial one, I'm afraid I don't follow. We had declared out initial hierarchy to be infinite; also, the definition of $V_{\alpha+1}$ and $V_{\omega + 1}$ are the same. Doesn't this mean we have achieved nothing because of the new infinite union? $\endgroup$
    – ankush981
    Commented Jul 24, 2014 at 6:49
  • $\begingroup$ Each $V_{\alpha+1}$ (for $\alpha\in\mathbb{N}$) is a finite union, while $V_{\omega+1}$ is an infinite union. So they are defined similarly, but are not the same. $\endgroup$
    – cws
    Commented Jul 24, 2014 at 14:49
  • $\begingroup$ Very nice! Thanks for all the help! :) $\endgroup$
    – ankush981
    Commented Jul 24, 2014 at 16:04

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