# Proof-check: If two sequences are $\varepsilon$-close, then the first one is bounded if and only if the second is bounded too

Definitions.

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.

• Do you want $\varepsilon$ or $\varepsilon/3$? – Michael Burr Jan 16 '18 at 14:35
• 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. – Michael Burr Jan 16 '18 at 14:37
• What about $b_m$ with $m<N$? – Michael Burr Jan 16 '18 at 14:37
• $| b_m| < \varepsilon + M_o$, for any $m$. Isn't it right? – DunhoClark Jan 16 '18 at 14:46
• No, $|b_m|<\varepsilon+M_0$ only for $m>N$. For $m\leq N$, you don't have the inequality $|b_m-a_m|<\varepsilon$. – 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 $(b_n)$.