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Let $(a_{n)n \in N}$ and $(b_{n)n \in N}$ be two convergent sequences of points in $\mathbb{R}^{d}$. Prove if $a_n \rightarrow a$ and $b_n \rightarrow b$, then $||a_n - b_n|| \rightarrow ||a-b||$ as $n \rightarrow \infty$

I know that if they $\lim_{n \to \infty} a_n \leq \lim_{n \to \infty} b_n$ for only sufficiently large values of n using the squeeze theorem. Would the same apply here or is there something else to be considered?

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If $a_n \to a$ and $b_n \to b$, it is clear that $a_n - b_n \to a-b$. To see this, consider the inequality $$\|a_n - b_n - a+b\| \le \|a_n - a\| + \|b_n - b\|$$ for all $n\in \mathbb N$. Hence, your question boils down to showing that if $c_n \to c$, then $\|c_n\| \to \|c\|$. The triangle inequality comes to our rescue once again. We have $$\|c_n\| \le \|c_n - c\| + \|c\| \tag{1}$$ and $$\|c\| \le \|c_n - c\| + \|c_n\| \tag{2}$$ Combining $(1)$ and $(2)$, we get $$|\|c_n\| - \|c\|| \le \|c_n - c\|$$ so that when $\|c_n - c\| \to 0$, we also have $\|c_n\| \to \|c\|$. Apply this result to $c_n = a_n - b_n$ to get the desired conclusion.

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  • $\begingroup$ Ohh I see, thank you so much! $\endgroup$
    – Axel
    Sep 23 at 13:37

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