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I understand the general method of the proof why a norm is a continuous function, for example https://math.stackexchange.com/a/265595/188401.

However I am struggling to comprehend why the smallness of the distance between $x$ and $y$ is also measured by the norm. So let $(X, \|\cdot \|)$ be normed space. Let $x, y \in X$. The norm is defined as a function $x \mapsto \|x\|$, so by definition the norm symbol $\| \|$ is assigned to the range, not the domain. So a priori why do we measure the distance $|x-y|$ in terms of the norm $\|x-y\|$, and not by something else?

Edit for clarity. This is a copy and paste of a part of the answer from the above link. We need to prove: $$\forall (x_n):\mathbb{N}\to X\ x_n\to x\implies \left\|x_n\right\|\to \left\|x\right\|$$

Let $\epsilon>0$ and $(x_{n})$ be an arbitrary sequence in $X$ that converges to $x \in X$. Then

$$\exists N\in \mathbb{N}:n\ge N\implies \left\|x_n-x\right\|<\epsilon$$

etc...

My question is: in the last line above, why do we measure distance between $x_{n}$ and $x$ in the norm? $x_{n}$ converges to some $x$, but they are just elements of $X$ before the norm function is applied...

Community
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Alex
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3 Answers3

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If you want to prove the continuity of the norm, you have to prove that if $x_n \to x$ with convergence in $X$ then, $\|x_n\| \to \|x\|$ with convergence in $\mathbb{R}$. So that, the question is to prove that $|\|x_n\|-\|x\| |\to 0$. But it is straightforward using the (reverse) triangular inequality for the norm.

anonymus
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First of all, the norm is a function whose codomain must also have some metric defined on it. That is, $\Vert \cdot\Vert: X\to (\mathbf F, \vert\cdot\vert)$.

Establishing the continuity of the norm means that when presented with $\epsilon \gt 0$, we must supply a $\delta \gt 0$ such that $$ \Vert x - x_0\Vert \lt \delta \implies \vert \Vert x\Vert - \Vert x_0\Vert\vert \lt \epsilon $$ The definition of a converging sequence $x_n \to x$ in a normed space $(X,\Vert\cdot\Vert)$ is that for each $\delta \gt 0$ there is some $N \in\Bbb N$ such that $n\in \Bbb N$ and $n\ge N \implies \Vert x_n - x\Vert \lt \delta$.

Alex Ortiz
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  • My question was essentially the reverse of your answer: I am asking why apriori the $\mathb{domain}$ of the norm has a a metric defined on it defined by the norm? I though the metric defined by the norm only applied to the codomain? @AOrtiz – Alex Sep 09 '16 at 17:18
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As a suggestion notes that $\left \| x \right \| \leq \left \| x-y \right \| + \left \| y \right \|$ and $\left \| y \right \| \leq \left \| x-y \right \| + \left \| x \right \|$, then$| \left \| x \right \| - \left \| y \right \| | \leq \left \| x-y \right \|$. Now just apply your considerations, and remember that a norm is a function $\left \| \cdot \right \| : X \longrightarrow [0,\infty)$.

user288972
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