How to show that the class of all sets of a particular cardinality ,say $h$ is not a set.

My argument:

I assume that I've shown the following lemma.

Lemma: If $X$ is an infinite set of cardinality $h$ , $a\ne x$ for every $x\in X$ then $\{a\}\cup X$ is of cardinality $h$ too i.e. adding new element to an infinite set doesn't change cardinality.

Let $h$ be an infinite cardinal number. Let $T_h=\{x:\text{x is a set of cardinality h}\}$. Suppose $T_h$ is a set.

Let $a$ be any set and consider any $t\in T_h$. Let $R=t\cup \{a\}$. Using the above lemma, $R\in T_h$.

So, $a\in R\subset \bigcup T_h$. As $T_h$ is a set then $\bigcup T_h$ is also a set. As $a$ was arbitrary set then we have, $a\in\bigcup T_h$ for every set $a$. Hence, the set $\bigcup T_h$ contains all sets which meas that the collection of all sets is a set, contradiction.

If $h$ is non-zero finite cardinal number. Define $T_h$ the same way as above. Let $a$ be any set, Let $A_1=a, A_{n+1}=A_n\times a$. Consider $A=\{A_1,A_2,...,A_h\}$. So, $A\in T_k$. Hence, $a=A_1\in \bigcup T_h=K$. So K is a set containing all sets $a$ since $a$ was arbitary,contradiction.

My main question is, Is there any flaws in my argument? If yes, point it out please.

If the argument is valid, Is there any standard proof to show the same fact?

  • 1
    $\begingroup$ We want non-zero cardinality. The argument looks OK, if we prove the lemma. Too complicated. Let $A$ be a non-empty set, and pick $a\in A$. For any set $x$ such that $\{x\}\ne a$, replace $a$ by $\{x\}$. Let $A_x$ be the resulting set. Easily cardinality does not change. The rest is as in your proof. $\endgroup$ – André Nicolas Jan 26 '15 at 18:04
  • $\begingroup$ @AndréNicolas, for the lemma, I think I can prove it easily by showing that there is an injection from $\mathbb{N}$ into any infinite set And then "hiding" the new element by moving images of natural numbers one step to the right $\endgroup$ – Fawzy Hegab Jan 26 '15 at 18:23
  • $\begingroup$ @AndréNicolas, But of course, your method is much more easier! and more intuitive. Thank you ... $\endgroup$ – Fawzy Hegab Jan 26 '15 at 18:35
  • $\begingroup$ You are welcome. Choosing a simpler way was also motivated by a wish not to have AC sneak in. $\endgroup$ – André Nicolas Jan 26 '15 at 19:00
  • $\begingroup$ @AndréNicolas, Yes, In fact, one of the things I wanted to avoid is not to use AC but To prove my lemma, I have to use it even implicitly. $\endgroup$ – Fawzy Hegab Jan 26 '15 at 19:06

Your argument is good for infinite $h$, but the claim is true for finite non-zero $h$ too. To see this you need to adjust your lemma to say that for any non-zero cardinal $h$ and for any set $a$, there is a set of cardinality $h$ that has $a$ as a member. To prove this note that if $x$ has $|x| = h > 0$ and $a \not\in x$, then $x = \{y\} \cup z$ for some $z$ with $y \not\in z$, but then $|\{a\} \cup z| = h$ and $a \in \{a\} \cup z$.

  • $\begingroup$ For finite $h$, I've updated the argument. Could you check it please? $\endgroup$ – Fawzy Hegab Jan 26 '15 at 18:23
  • $\begingroup$ Your comment on finite cardinals is similar to the one made by André Nicolas. In fact Showing that every set $a$ is an element in some set of cardinality $h$ for every $h$ was the thing I was trying to prove using the lemma. I didn't think of just "replacing one element!" although It's much more easier... $\endgroup$ – Fawzy Hegab Jan 26 '15 at 18:34
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    $\begingroup$ You don't need separate cases for finite and infinite $h$. The argument in my answer (which is essentially the same as the one in AndreNicolas's comment) applies to finite and infinite $h$. $\endgroup$ – Rob Arthan Jan 26 '15 at 18:43
  • $\begingroup$ Yes, I see. what I was trying to say is that, even without editing the lemma, We can cover the other case of finite cardinals. In fact, my argument for finite cardinals- if you read it- is no more than finding an explicit set of cardinality $h$ which contains $a$. $\endgroup$ – Fawzy Hegab Jan 26 '15 at 18:47

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