Bounded sequences. Let $(a_n)_{n=0}^\infty$ be a sequence. It is bounded iff there is an $M_0$ such that $|a_n| \leq M_0$.

Eventually $\varepsilon$-close. We say two sequences are eventually $\varepsilon$-close iff, for some $\varepsilon>0$, there exists an $M_1$ such that, for any $n, m \geq M_1$, $|a_m - a_n| < \varepsilon$.

Proposition. Let $\varepsilon > 0$. Show that if $(a_m)_{m=0}^\infty$ and $(b_m)_{m=0}^\infty$ are eventually $\varepsilon$-close, then $(a_m)_{m=0}^\infty$ is bounded if and only if $(b_m)_{m=0}^\infty$ is bounded.

Proof. Assume $(a_m)_{m=0}^\infty$ and $(b_m)_{m=0}^\infty$ are eventually $\varepsilon$-close for some $\varepsilon$. So there exists an $N$ such that, for any $m \geq N$, $|b_m - a_m | < \varepsilon$.

Suppose $(a_m)_{m=0}^\infty$ is bounded. So there exists an $M_0$ such that $|a_m| \leq M_0$.

Then $$|b_m - a_m | + | a_m| < \varepsilon + M_0$$

By the triangle inequality theorem,

$$ |b_m - a_m + a_m | =|b_m| < \varepsilon + M_0 $$

This implies that $(b_m)_{m=0}^\infty$ is bounded by an $M_1$ such that $M_1 = max(b_1, b_2, b_3, ..., \varepsilon + M_0$).

We could show that if $(b_m)_{m=0}^\infty$ is bounded, $(a_m)_{m=0}^\infty$ is bounded too by a reciprocal argument.

  • $\begingroup$ Do you want $\varepsilon$ or $\varepsilon/3$? $\endgroup$ – Michael Burr Jan 16 '18 at 14:35
  • $\begingroup$ You really should have an inequality of the form $|b_m|\leq$something. You have all the details, but you don't reach this statement. $\endgroup$ – Michael Burr Jan 16 '18 at 14:37
  • $\begingroup$ What about $b_m$ with $m<N$? $\endgroup$ – Michael Burr Jan 16 '18 at 14:37
  • $\begingroup$ $| b_m| < \varepsilon + M_o$, for any $m$. Isn't it right? $\endgroup$ – DunhoClark Jan 16 '18 at 14:46
  • $\begingroup$ No, $|b_m|<\varepsilon+M_0$ only for $m>N$. For $m\leq N$, you don't have the inequality $|b_m-a_m|<\varepsilon$. $\endgroup$ – Michael Burr Jan 16 '18 at 15:45

All the pieces are there, here's how to put them together. For $m \geq N$, you start with: $$ |b_m|=|b_m-a_m+a_m|. $$ Then you use the triangle inequality: $$ |b_m - a_m + a_m| \leq |b_m - a_m| + |a_m| < \varepsilon + M_0. $$ By transitivity, this gives you: $$ |b_m| < \varepsilon + M_0. $$ And as you noted, by taking the max of $\{ |b_1|, \ldots, |b_{N-1}|, \varepsilon+M_0 \}$ you get a bound for $(b_n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.