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Show that ${x_{k}}\rightarrow x$ in $R^n$ if and only if ${x_{k}}\cdot y\rightarrow x\cdot y$ for all $y\in R^n $

I think the following:

Given the ${x_{k}}\rightarrow x$ this is equal to $\lim\limits_{k\rightarrow \infty}x_{k}=x$, then given a $y\in R^n$, we have to:
\begin{eqnarray*} {x_{k}}\rightarrow x &\Leftrightarrow& \lim\limits_{k\rightarrow \infty}x_{k}=x\\ &\Leftrightarrow& \lim\limits_{k\rightarrow \infty}x_{k} \cdot y = x \cdot y\\ &\Leftrightarrow& {x_{k}}\cdot y \rightarrow x \cdot y \end{eqnarray*}

Any suggestions and comments are very helpful, thanks in advance.

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    $\begingroup$ Try to write the proof by the definition of convergent sequence this would be much better $\endgroup$
    – 00GB
    May 19, 2021 at 18:00

1 Answer 1

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Hint 1: Remember that a sequence in the metric space $\mathbb{R}^n$ (with the usual metric) converges to a point in the space if and only if the component sequences converge.

Hint 2: for every $i= 1, \dots , n $ make $y = e_i$, the vector whose k-entry is 1 whereas the other entries are 0.

Edit: By a component sequence of $\{x_k\}$ I mean the sequence whose $k$th term is the $i$th coordinate of the point $x_k$ from the original sequence (where $i$ is some fixed integer of the set $\{1, \dots n\}$). Thus, since you are dealing with a sequence of points on $\mathbb{R}^n$ (points described with n coordinates), you can construct $n$ of those sequences.

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