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Here is a proof for the statement that the set of natural numbers N is not equivalent to (0,1):

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Firstly, I do not understand that why this proof shows that the set of natural numbers N is not equivalent to (0,1). Let us say the proof does show that there is a real number r in (0,1) such that r does not equal to $a_n$ for every n. But how does that relate to the statement that we need?

Secondly, in the proof, I do not understand why $I_{n+1} ⊂ I_{n}$ follows from the selection of $I_n$.

I would really appreciate if someone could give me a hand! Thanks so much.

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    Well the proof is too hastily or loosely written. By "closed subinterval", we finally rectify "closed subinterval with non-empty interior" and this works. – 2017-02-27

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Ok. What is the proof trying to setup by considering such an arbitrary sequence $(a_n)$? Why in the world, are we talking about sequences, even when all we need to show is something completely unrelated?

... Think about it for a while. Are these really unrelated things?

Start from the beginning. We need to show that $\mathbb{N}$ and $(0,1)$ are not equivalent (I assume you're talking about their cardinalities). Let's say, they were. Then, next?

Definition: Two non-empty sets are said to have same cardinalities, if there exists a bijection between the two sets.

So let's say there is a bijection from $(0,1)$ to $\mathbb{N}$.

Here comes something interesting and "related". If we have a bijection from a set to $\mathbb{N}$, we can represent each element of that set with a unique natural number. What that means is we can represent the elements of this set as $a_1, a_2, \ldots$, which is implicitly representing the bijection itself (How?)

So, the proof now carries on with the rest of the argument. And what it tries to present is that, even if we had a bijection then there is an element in $(0,1)$ which is not mapped to any element of $\mathbb{N}$ (doesn't appears in the sequence $(a_n)$ anywhere), but that shouldn't be the case as we had clearly set up a mapping between the two, that is to say, each element from $(0,1)$ was enumerated i.e. appeared in our sequence somewhere or the other.

For the next question, why is $I_n \subset I_{n+1}$. This is by our construction.

Let's say, you had $0.958 = a_1, 0.123 = a_2, 0.456 = a_3, \ldots$ (doesn't matter, take up anything you like), then choose $I_1 = [0.1,0.9], I_2 = [0.2,0.9], I_3 = [0.2,0.3]$. Do you see the pattern?

Don't get involved with so many symbols, replace them with whatever numbers you like, for your own understanding.

Good luck.

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    Just to confirm: The sequence takes value from (0,1), right? – 2017-02-27
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    Yes! Indeed, each and every value in $(0,1)$ as per our assumption. – 2017-02-27
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    Thanks so much. That is what confused me. I thought it was just an arbitrary sequence of real numbers. Very Clear Explanations! – 2017-02-27
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    Pleasure was all mine! – 2017-02-27
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I assume by equivalent you mean of the same size or "cardinality", i.e. equipollent. The point is that if a set is equipollent to $\mathbb{N}$ there exists a bijection between the two, from which we obtain a sequence that contains all elements of the set. But the argument shows no such sequence exists (i.e. a sequence within the set doesn't exhaust it), a contradiction.

As for $I_{n+1}\subseteq I_n$, that follows from the fact that $I_{n+1}$ is defined as a certain subset of $I_n$.

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    For my second question, does that just come from the definition or is that related to the selection of $I_n$? – 2017-02-27
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    @ButterMath A subinterval of an interval is also a subset thereof, for the simple reason nothing can be in the subinterval unless it's in the interval as well. – 2017-02-27
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The proof shows that for any sequence of real numbers there is always another real number within the interval, which is not in the sequence. It's equivalent to Cantor's diagonal argument, since a sequence of real numbers is countable but an interval of real numbers always contains uncountably many.

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    It seems to me that Cantor's argument shows that there is no surjection from the set N to (0,1). Why does the argument that I provide not involve N at all? – 2017-02-27
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    Is that because the enumerated natural numbers (N) also form a sequence of real numbers? – 2017-02-27
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    A sequence (where the index set is $\mathbb{N}$) is by definition a map from $\mathbb{N}$ to the set in which the sequence takes values, in this case $(0, 1)$. – 2017-02-27
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    @Bib-lost I got it now. Thanks so much. – 2017-02-27