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

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

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

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

Is it enough to say that if there was a largest element in $\displaystyle S_\Omega$ 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 $\displaystyle \alpha\in S_\Omega$, the subset $\displaystyle \{x|\alpha<x\}$ is uncountable.

Can I say: Let $\displaystyle S=\{x|\alpha<x\}$ and let $\displaystyle T=\{x|x\leqslant\alpha\}$ be a countable set. Then $\displaystyle S_\Omega = S\cup T$. Since $\displaystyle S_\Omega$ 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 $\displaystyle \{x|\alpha<x\}$ 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 $\displaystyle S_\Omega$ be the minimal uncountable well-ordered set.

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

Is it enough to say that if there was a largest element in $\displaystyle S_\Omega$ 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 $\displaystyle \alpha\in S_\Omega$, the subset $\displaystyle \{x|\alpha<x\}$ is uncountable.

Can I say: Let $\displaystyle S=\{x|\alpha<x\}$ and let $\displaystyle T=\{x|x\leqslant\alpha\}$ be a countable set. Then $\displaystyle S_\Omega = S\cup T$. Since $\displaystyle S_\Omega$ 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 $\displaystyle \{x|\alpha<x\}$ 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 $\displaystyle S_\Omega$ would not be minimal.