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I know that real number is uncountable,the same as the closed interval [0,1].However,I don't know what is wrong with my method of counting numbers in [0,1]:

0.1,0.2…0.9

0.01,0.02…0.99

0.01,0.02,…0.999

....

Can anyone tell me what'wrong with it?Thanks.

2 Answers 2

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You are merely counting rational numbers, not all real numbers between $0$ and $1$. The problem with the real numbers is that from any list of real numbers, using Cantor's diagonal argument, one can show that there exists a real number that is not in that list. This is why the real numbers are uncountable.

Have fun reading this : http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

Hope that helps,

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    and even not all rationals, just with denominator $2^a5^b$.2012-10-20
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    I know what you mean thanks.2012-10-20
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    Now I know what is called countable set.That means that you devise a method to count with which every element can be placed in a particular position n.2012-10-20
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    @89085731 : Yes, exactly. The "method of counting" would be the bijection between your countable set $S$ and the set of positive integers $\mathbb N$, and if you count the elements by $S = \{x_1, \dots, x_n, \dots \}$, then the bijection is $i \mapsto x_i$. By the way, if the answer I gave you answers your question you should click on the check button under the downvote button.2012-10-20
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When does your enumeration reach pi? Or Sqrt(2)? Or even 1/13? It doesn't.

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    $\pi \notin [0,1]$, nor does $\sqrt 2$. But you're right about $\frac 1{13}$.2012-10-20
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    @PatrickDaSilva: true, but he could ask about $\pi -3$ and $\sqrt 2 -1$2012-10-20
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    Sure, I was just mentioning.2012-10-20