TITLE: Why is the collection of all algebraic extensions of F not a set?
QUESTION [11 upvotes]: When proving that every field has an algebraic closure, you have to be careful. In this article https://proofwiki.org/wiki/Field_has_Algebraic_Closure, and as I have been told on this site, if we have a field F. The "collection of all algebraic extensions of F" is not a set.
Is there a simple way to explain why this is not a set, and we can not apply zorns lemma on it? Or do you need a lot of reading in deep set-theory and logic to understand this? I have seen the russel paradox, but that is basically how much I know about this.
What also is very confusing is that in real analysis we have that "the space of continuous functions on [0,1] is a vector space". So there is a set of continuous functions? This doesn't sound any more mysterious than "all algebraic extensions of a given field F", however one of them gives rise to a set, and one doesn't?

REPLY [10 votes]: Here's a reformulation of the argument that avoids the notion of cardinality.  Suppose $F$ is not algebraically closed, let $K$ be some algebraic extension of it, and let $a$ be any element of $K$ that isn't in $F$.  Then, for any entity $x\notin K$, we can produce another algebraic extension of $F$, isomorphic to $K$, by replacing the element $a$ in $K$ with $x$.  So every $x\notin K$ is in an algebraic extension of $F$, and so is every $x\in K$ (because $K$ itself is an algebraic extension of $F$).  If there were a set of all algebraic extensions of $F$, then the union of this set of fields would be a set (by the axiom of union) and yet would contain everything.  That contradicts the theorem (of standard set theory) that no set can contain everything.