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This question is related to Continuity of vector space operations in a normed space. To be precise, it clarifies what continuity means when proving the link's question. It is a follow-up.

Show that in a normed space $X$, vector addition and scalar multiplication are continuous operations with respect to the norm; that is, the mappings defined by

$(x,y) \mapsto x+y$

$(\alpha,x)\mapsto \alpha x$

are continuous.

The following pdf argues that one way to prove continuity is use a sequential definition of continuity. That is, we seek to show for vector addition that whenever $x_n \rightarrow x$ and $y \rightarrow y$ then $x_n + y_n \rightarrow x + y$. For scalar multiplication, a similar variant is to be shown.

I "contest" that we really need to show that whenever $(x_n, y_n) \rightarrow (x,y)$, then $x_n+y_n \rightarrow x + y$. And here is the subsequent ambiguity: what does it mean for

\begin{equation}(x_n, y_n) \rightarrow (x,y) \qquad (\alpha)\label{eqn:a}\end{equation}

If were considering vector addition as a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, then to show $(\alpha)$, one would have to show that $\sqrt{(x_n - x)^2 + (y_n - y)^2}$ can be made arbitrarily small. With the case of an arbitrary norm, I do not show understand how one might show $(\alpha)$.

How might one handle such an arbitrary norm, or why is my "contest" not legitimate.

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As every norm $||\cdot||$ induces a metric $d:V\times V\rightarrow\mathbb{R}$ by $d(x,y)=||x-y||$, the convergence can be proven by checking if both $d(x_n, x)\rightarrow0$ and $d(y_n, y)\rightarrow0$.

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  • $\begingroup$ If I have understood you correctly, then you have argued that $d(x_n ,x ) \rightarrow 0$ and $d(y_n, y) \rightarrow 0$ implies $\alpha$, but what does $\alpha$ itself mean in the context of the norm $| | \cdot | |$? $\endgroup$ – Muno Jun 4 '17 at 19:24
  • $\begingroup$ First of all, convergence in the product space is implied by convergence of the individual components. Secondly, in a metric space (we turned your normed space into a metric one by the formula i gave) the convergence of a sequence is implied by $d(x_n,x)\rightarrow0$. $\endgroup$ – NDewolf Jun 4 '17 at 19:36
  • $\begingroup$ So $\alpha$ can be understood easily by viewing the normed space as a metric space. $\endgroup$ – NDewolf Jun 4 '17 at 19:37

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