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Question: Show that the interval $⟨2, 5⟩ ⊆ ℝ⁺$ is an uncountable set.\

To show that the interval $ \langle 2, 5 \rangle \subseteq \mathbb{R}^+ $ is an uncountable set, we can use Cantor's diagonal argument.

Assume for contradiction that $ \langle 2, 5 \rangle $ is countable. Then we can list its elements as follows:

$ x_1, x_2, x_3, \ldots $

where each $ x_i $ is a real number in the interval $ \langle 2, 5 \rangle $.

Now, let's construct a number $ y $ such that its $ i $th decimal digit is different from the $ i $th decimal digit of $ x_i $.

Then, $ y $ is a real number in the interval $ \langle 2, 5 \rangle $, but it is not in our list because it differs from each $ x_i $ at least in one decimal place. This contradicts the assumption that we could list all the elements of $ \langle 2, 5 \rangle $.

Therefore, $ \langle 2, 5 \rangle $ is uncountable.

Is this correct approach or am I doing something wrong?

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    $\begingroup$ It's correct if you can show construction of $y$ s.t. it's in $(2, 5)$. $\endgroup$
    – mihaild
    Commented Apr 7 at 17:20
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    $\begingroup$ Another approach might be to find a bijection between $(2,5)$ and $\mathbb R$ $\endgroup$
    – J. W. Tanner
    Commented Apr 7 at 17:37
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    $\begingroup$ It's correct. As @J.W.Tanner pointed out, you can also show that there exists a bijection $f : (2,5)\to\mathbb R$, even though in that case you'd be assuming that $\mathbb R$ is uncountable in the first place. Anyways, a simple bijection could be $f(x) = \tan\left(\dfrac\pi3x-2+\dfrac\pi2\right)$, the domain being of course $(2,5)$. $\endgroup$
    – NtLake
    Commented Apr 7 at 20:58

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