# Thread: Is this enough to solve minimal uncountable set problem?

1. ## Is this enough to solve minimal uncountable set problem?

Let $S_\Omega$ be the minimal uncountable well-ordered set.

(a) show that $S_\Omega$ has no largest element.

Is it enough to say that if there was a largest element in $S_\Omega$ and we were to remove it, we would still be left with an uncountable set and an ordinal set which will then contradict "minimality".

(b) Show that for every $\alpha\in S_\Omega$, the subset $\{x|\alpha is uncountable.

Can I say: Let $S=\{x|\alpha and let $T=\{x|x\leqslant\alpha\}$ be a countable set. Then $S_\Omega = S\cup T$. Since $S_\Omega$ is uncountable (defined in he beg. of the problem), then either S or T has to be uncountable and seeing as T is already defined as a countable set it follows that S must not be countable. Thus, the set $\{x|\alpha is uncountable.

Note: I apologize in advance for my latex. The S_Omega is supposed to be S (subscript Cap Omega).

2. Originally Posted by tn11631
Let $S_\Omega$ be the minimal uncountable well-ordered set.

(a) show that $S_\Omega$ has no largest element.

Is it enough to say that if there was a largest element in $S_\Omega$ and we were to remove it, we would still be left with an uncountable set and an ordinal set which will then contradict "minimality".
Yes, that's fine.

Originally Posted by tn11631
(b) Show that for every $\alpha\in S_\Omega$, the subset $\{x|\alpha is uncountable.

Can I say: Let $S=\{x|\alpha and let $T=\{x|x\leqslant\alpha\}$ be a countable set. Then $S_\Omega = S\cup T$. Since $S_\Omega$ is uncountable (defined in he beg. of the problem), then either S or T has to be uncountable and seeing as T is already defined as a countable set it follows that S must not be countable. Thus, the set $\{x|\alpha is uncountable.
The way you have written this, it looks as though you are assuming the result that you should be trying to prove. If T is countable then it follows that S is uncountable, as you have correctly shown. But you can't just assume that T is countable, you need to prove it. In fact, that is not hard, because if T was uncountable then $S_\Omega$ would not be minimal.