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I need to prove that $\lim_{n \to \infty} \sup \{2^k : 2^k \leq n\} = \infty$.

I know that the supreme exists, the set is non-empty ($\forall n \geq 1$ : $2^{-1} \in \{2^k : 2^k \leq n\}$). I also know that the set is bounded by aboven (by $n$). But how can I correctly say that this limit goes to infinity?

Michael Hardy
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user54297
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    If you call the sets $E_n$ then $\sup E_{2^n} = 2^n$ which gets very big. –  Mar 31 '14 at 13:36
  • Not every question involving sets is about set theory. – Asaf Karagila Mar 31 '14 at 13:54
  • @user54297 : Don't write \mbox{ sup }. I changed it to \sup. That not only prevents italicization but also results in proper spacing in things like $a\sup b$ but also affects format when things like $\displaystyle\left(\sup_{a\in S} \cdots\right)$ appear in a displayed, as opposed to inline, setting. It is standard usage. ${}\qquad{}$ – Michael Hardy Mar 31 '14 at 14:18

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Let $M_n := \{2^k : 2^k \le n\}$. For $n_1 \le n_2$ you get $\sup M_{n_1} \le \sup M_{n_2}$. Therefore you only need to look at a subsequence: Let‘s take $\sup M_{2^l}$ which is equal to $l$. So for $l \rightarrow \infty$ you get $\sup M_{2^l} \rightarrow \infty$.

Keba
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Let $M_n = \{2^k\mid 2^k\le n\}$.

We then have $\frac12n < \sup M_n$, because there is a number of the form $2^k$ in $(\frac 12 n, n]$.

Therefore, from the definition of limit, if you have $K$ and want to find $N$ such that $\sup M_n\ge K$ for all $n\ge N$, you can just take $N=2K$.

hmakholm left over Monica
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