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I want to show the set of all irrational number in $(0,1)$ is uncountable.

I know that the set of all irrational number is uncountable and i know that the superset of uncountable set is uncountable but i don't know how can i use this? Please help me or give me some other way!

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  • $\begingroup$ Can you prove that $(0,1)$ is uncountable? $\endgroup$ – Alex Becker Sep 9 '12 at 5:20
  • $\begingroup$ Yes because $(0,1)$ is equivalent to $[0,1]$ and $[0,1]$ is uncountable so $(0,1)$ is uncountable. $\endgroup$ – Kns Sep 9 '12 at 5:23
  • $\begingroup$ And the rationals are countable, and the union of two countable sets is countable, so if the irrationals are countable as well then... $\endgroup$ – Alex Becker Sep 9 '12 at 5:28
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Hint: With $A$ uncountable, $B$ countable, and $B\subset A$, prove that $A-B$ is uncountable. What would go wrong if it were countable? Apply this result to $(0,1)$ and the rationals within it.

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From your comment, I assume that you know that $(0,1)$ is uncountable. Let $I$ denote the set of irrational numbers. Then $(0,1)$ is the disjoint union of $(I \cap (0,1))$ and $(\mathbb{Q} \cap (0,1))$. Since $\mathbb{Q}$ is countable, its subset $(\mathbb{Q} \cap (0,1)$ is also countable. If $(I \cap (0,1))$ was countable, then $(0,1)$ would be the union of two countable sets and hence $(0,1)$ would be countable. Contradiction. Thus $I \cap (0,1)$ is uncountable. $I \cap (0,1)$ are precisely the irrationals in $(0,1)$.

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Alternatively, $$ f(x)=\begin{cases}2-\frac{1}{x} & x<1/2 \\[1ex] 0 & x =1/2 \\[1ex] \frac{1}{1-x}-2 & x > 1/2 \end{cases} $$ maps $(0,1)$ bijectively to $\mathbb R$ such that rationals map to rationals and irrationals to irrationals.

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