Suppose I have two sequences, $t=(t_1,t_2,... )$ in $[0,1]$ and $y=(y_1,y_2,...)$ in $\mathbb{R}$. Is it possible to construct a continuous function $f:[0,1]\longrightarrow \mathbb{R}$ such that $f(t_i)=y_i$ $\forall i \in \mathbb{N}$? If it is possible, how can it be done? Thanks a lot in advance!
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Newton Polynomial? – Kaster Sep 04 '14 at 23:44
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Newton polynomial does this when we don't have sequences, but finite set of points instead. – Vokram8 Sep 04 '14 at 23:46
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My question is exactly if it would be posible to find a continuous function f such that f(r_i)=i, where r_i is the i-th rational. – Vokram8 Sep 04 '14 at 23:52
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It is not possible! It follows from the definition of continuity. If $x,y$ are two different rational numbers then $|f(x)-f(y)| \geq 1$ so $f$ cannot be continious at any point. – Winther Sep 04 '14 at 23:53
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@Vokram8 oh, my bad. Misunderstood the part with finite and infinite but countable set. – Kaster Sep 05 '14 at 04:19
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In general you cannot do this. For example, let $t = (0, 1/2, 1/4, 1/8, \ldots)$, and $y = (0, 1, 1, 1, \ldots)$. Such a continuous function would necessarily take the value $0$ at $x = 0$, but take the value $1$ at points arbitrarily close to $0$.
Edit: In addition, if $y$ contains arbitrarily large real numbers, then such a continuous function will not exist, since the image of this function should be compact.
