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This seems a very statement to be proven, but I'm stuck. I'm trying the contradiction technique:

Take $i>j$. Then, we know that $K(i) > K(j)$. Suppose for the sake of obtaining a contradiction that $K(i) < i$. I can't follow from here. Any hints?

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    Proof by induction. – 2017-01-29
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    The function K(i) = i - 1 is a strictly increasing function and K(i) = i - 1 < i for all i. So the statement is false without some other assumption on K. For example, does K only take values in the natural numbers? – 2017-01-29
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    Yes, that's right! I will edit it. It only takes values on natural numbers! – 2017-01-29

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Let $K:\mathbb{N} \to \mathbb{N}$ be a strictly increasing function, that is, $K(n) \geq K(n-1) + 1$, and $K(1) \geq 1$.

Claim: $$K(i) \geq i \quad \forall i \in \mathbb{N}$$

Base Step: $K(1) \geq 1$.

Induction Step:

$$K(i+1) \geq K(i) + 1 \geq i + 1. $$

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    @kobe better ? :P – 2017-01-29
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    It's all good. :) – 2017-01-29