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@KingGeorge Can you take a look?

The thing is we dont touch on Cantor's Diagonalization. We are learning abt limts and covergence

So your teacher is looking for something that's not the diagonalization?

Yess

Gonna think abt it toooo

Well actually it gives us the method we should be using

OK whatever its 3:30 am. I cant think anymore. i will have to put something together tomorrow

For (a), perhaps a proof by induction will work.
Base: n=1. Let f(1)=\(x_1\in[0,1]\). Then you can choose \(a_1\in[0,1]\) such that \(a_1\) is minimal and less than \(x_1\). Similarly, let \(b_1\in[0,1]\) such that \(b_1\) is maximal and less than \(x_1\). If \(x_1=0\), choose any \(a_1,b_1\in(0,1]\). Clearly, \(x_1\notin[a_1,b_1]\).
Now assume true up to some \(k\in \mathbb{N}\). Then we have two cases.
Case 1: \(x_{k+1}\notin[a_k,b_k]\). Here, just choose \(a_{k+1}>a_k\) and \(b_{k+1}

Then, we can see that A is in \([a_n,b_n]\) for all \(n\in\mathbb{N}\) since each interval is contained within the previous one, and since \(\{a_n\}\) is a strictly increasing sequence. If \(A\notin[a_n,b_n]\) for some n, then since the sequence is increasing, it must be that \(A>b_k\) for some \(k\). However, \(a_k

That was the onlt part that was confusing me

No wait. Let me read over that more carefully before I continue.

That make more sense?

Ohhhh ok get it

Yaaaaaa

So we created the sequence \(\{a_k\}\) as a strictly increasing sequence (at least with the new revisions I just gave a few posts above). Thus, \(A>a_k\) for any \(k\in\mathbb{N}\). Therefore, we now suppose that \(A\notin[a_k,b_k]\) for some \(k\in\mathbb{N}\). So \(A>b_k\). However, since \(\{a_k\}\) converges to \(A\), we can choose some \(a_j\) such that \(b_k

OHHHH I GET IT NOWWW

Thanks KG :)

Its really clear and and to the point. You always have this sleek way of proving

Thanks :)