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Describe why norms are continuous function by mathematical symbols.

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A function $f$ from a metric space to a metric space is continuous if for all $x$ in the domain, for all $\varepsilon>0$, the exists $\delta>0$ such that for all points $y$ in the domain, if the distance from $x$ to $y$ is less than $\delta$, then the distance from $f(x)$ to $f(y)$ is less than $\varepsilon$.

If $f$ is a norm, then it maps a vector space into $\mathbb R$, and the distance from $x$ to $y$ is $f(x-y)$.

In this case it suffices to take $\delta=\varepsilon$, for the following reason. Suppose the distance from $x$ to $y$ is less than $\delta=\varepsilon$. Then $f(x-y)=f(y-x)<\varepsilon$ (where the equality follows from the definition of “norm”). Now recall that norms satisfy a triangle inequality:

$$

f(x) \le f(y) + f(x-y)

$$

$$

f(y) \le f(x) + f(y-x)

$$

So

$$

f(y)-f(x) \le f(y-x)<\varepsilon\text{ and }f(x)-f(y) \le f(x-y)<\varepsilon,

$$

so

$$

|f(x)-f(y)|<\varepsilon,

$$

i.e.

$$

\Big(\text{distance from $f(x)$ to $f(y)$}\Big) <\varepsilon.

$$

Let $(X,\left\|\cdot\right\|)$ be a normed space. We need to prove that:

$$\forall (x_n):\mathbb{N}\to X\ x_n\to x\implies \left\|x_n\right\|\to \left\|x\right\|$$

Let $\epsilon>0$ and $(x_n)$ be an arbitrary sequence in $X$ that converges to $x\in X$. Then,

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

But

$$ \left| \left\|x_n\right\|- \left\|x\right\|\right|\le \left\|x_n-x\right\|$$

by the triangle inequality. Thus,

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

and we are done!

To keep it short and straight to the point: the norm of the normed space $(X,\|\cdot\|)$ is a continuous function because the topology you (usually) consider on $X$ is the smallest topology in which $\|\cdot\|$ is continuous. So it is continuous because we want it to be continuous.

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