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A sequence {$x_n$} is said to be eventually bounded if there exists a positive integer $N$ and a positive number $M$ such that |$x_n$|< M for all $n>N$. Prove that a sequence is bounded if and only if it is eventually bounded.

Progress: Since this is an iff statement, I was able to prove the first one.

(→) Suppose sequence {$x_n$} is eventually bounded. We will show that {$x_n$} is bounded. Since after $N$ terms the sequence is bounded, we will consider the finite set {$x_1,x_2,…,x_N$} and $x_{N+i}$ for $i\in N$ (which we know is bounded). Let $U_1$ and $L_1$ be the upper and lower bound of {$x_1,x_2,…,x_N$} respectively and $U_2$ and $L_2$ be the upper and lower bound of $x_{N+i}$ for $i\in N$ respectively. Moreover, let $U$ be the upper bound of {$x_n$} then $U=max⁡(U_1,U_2)$ and similarly, let $L$ be the lower bound of {$x_n$} then $L=max⁡(L_1,L_2)$. So, {$x_n$ } has both upper and lower bound then {$x_n$} is bounded.

(←)Suppose sequence {$x_n$} is bounded. We will show that {$x_n$} is eventually bounded. "I am stuck here. Please help! Any hint or help will be much appreciated."

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  • $\begingroup$ Just take $N=1$. $\endgroup$ – TonyK Apr 7 at 15:24
  • $\begingroup$ Well, if it is bounded, you can take any bound for $M$, and any value for $N$. He who can do the more can do the least. $\endgroup$ – Bernard Apr 7 at 15:25
  • $\begingroup$ What do you mean? Can you clarify further? $\endgroup$ – mathhunterx Apr 7 at 15:25
  • $\begingroup$ If the sequence is bounded, it is a fortiori eventually bounded. $\endgroup$ – Bernard Apr 7 at 15:28
  • $\begingroup$ So you mean no need for formal proof? $\endgroup$ – mathhunterx Apr 7 at 15:37

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