Why is the collection of all algebraic extensions of F not a set? 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?
 A: If an algebraic closure $K$ of $F$ has cardinality $\kappa$, then every set of cardinality $\kappa-|F|$ can be unioned with $F$ to give an algebraic closure, by just defining the addition and multiplication by transport via the bijection with $K$. Thus, the class of sets of cardinality $\kappa-|F|$ (which we'll assume is nonzero, i.e., $F$ is not closed already) injects into the class of algebraic closures of $F$. But the class of all sets of a given nonzero cardinality is not a set, and so the class of algebraic closures is not a set either.
Continuous functions $[0, 1]\to\mathbb R$ form a set because $[0, 1]\times\mathbb R$ is a set, and the set of continuous functions $[0, 1]\to\mathbb R$ form a subset of the power set of $[0, 1]\times\mathbb R$.
A: Suppose you have a proper algebraic extension $K$ of $F$ (so $F$ is not algebraically closed).
If $X$ is any set disjoint from $F$ with $|X|=|K\setminus F|$, and $f_X\colon K\setminus F\to X$ is a bijection, then we can build a bijection $g_X\colon K\to F\cup X$ so that
$$
g_X(a)=\begin{cases}
a & \text{if $a\in F$}\\
f_X(a) & \text{if $a\in K\setminus F$}
\end{cases}
$$
and transport the field structure from $K$ to $F\cup X$ so that $g_X$ is a field isomorphism. Thus $F\cup X$ is an algebraic extension of $F$.
Since the class of sets equipotent to $K\setminus F$ and disjoint from $F$ is not a set, we're done.
The situation is completely different in the case of the continuous functions from $[0,1]$ to $\mathbb{R}$. But there is no “set” of all continuous functions to $\mathbb{R}$ with an arbitrary metric space as domain.
A: 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.
