Equivalent norms induces equivalent metric and equivalent metrics induces the same topology.

I have the definition on hand for what it means for metric, norms and normed space to be equivalent. What is unclear is the language "induced" used.

Any clarification to shed light is appreciated.

Thanks in advance.



Let V be a vector space over the field F of reals $\mathbb{R}$.

All norms $\left \| \cdot \right \|: V\rightarrow \mathbb{F}$ satisfies the homogeneity condition

$\left \| \alpha \vec{v} \right \|=\left | \alpha \right |\cdot \left \| \vec{v} \right \|: \forall \alpha \in \mathbb{F}, \vec{v} \in V$

Define the metric

$d: V \times V\rightarrow \mathbb{F} \left ( \vec{x},\vec{y} \right )\mapsto \left \| \vec{x}-\vec{y} \right \|$ So the metric defined by the Norm is such that

$d\left ( \alpha \vec{x} ,\alpha \vec{y} \right )=\left | \alpha \right |\cdot \left \| \vec{x}-\vec{y} \right \|=\left | \alpha \right |d\left ( \vec{x},\vec{y} \right )$

Suppose $\left \| \cdot \right \|, \left | \cdot \right |$ are both equivalent Norms. Then, we have $\alpha_{1}d\left ( \vec{x},\vec{y} \right ) \leq e\left ( \vec{x},\vec{y} \right ) \leq \alpha_{2}d\left ( \vec{x},\vec{y} \right )$

Recall: A topology induced by a metric. Let $\left ( X,d \right )$ be a metric space and $\tau$ the collection of arbitrary union of open balls in X. $\tau$ is called the metric topology induced by the metric d.

By definition, this is the metric topology.

In the metric space $\left ( V,\left \| \cdot \right \| \right )$ over the field $\mathbb{F} of reals \mathbb{R}$, the open balls are

$B_{\epsilon }\left ( \vec{a} \right )=\left \{ \vec{x} \in V : d\left ( \vec{x},\vec{a} \right ) = \left | \vec{x}-\vec{a} < \epsilon \right |\right \} $ for $\epsilon >0, \vec{a} \in V$.

$\tau$ is a topology on V called the standard topology.

  • $\begingroup$ "Induced" means just the corresponding topology to the open balls. $\endgroup$
    – Piquito
    Commented Nov 2, 2016 at 11:29
  • $\begingroup$ Nitpick: you already have a special notation for $\mathbb R$ - namely, the symbol $\mathbb R$. Don't go saying things like "let $\mathbb F = \mathbb R$ be the field of real numbers". $\endgroup$
    – Jack M
    Commented Nov 2, 2016 at 12:21
  • $\begingroup$ Noted @JackM But I'll like feedback on my attempt. I feel as though my understanding of the question is still rather fragmented and this would, hopefully, be reflected in my attempt so that I may correct any errors. $\endgroup$ Commented Nov 2, 2016 at 12:44

2 Answers 2


Define the metric induced by a norm on a vector space $V$ as $$ d(x,y) = || x - y||. $$ Consider the case of the real numbers with the norm as the absolute value of a number. We indeed have the metric as the absolute difference in this case.

The topology induced by a metric is the topology generated of all open balls in that space, i.e for any point in the space $X$, the neighborhood of radius $\varepsilon$ around that point is an open set. Again, compare with the standard topology on the reals with the usual metric.

  • $\begingroup$ I have made an attempt in the OP. Would you mind taking a look? $\endgroup$ Commented Nov 2, 2016 at 12:19

The topology induced (generated) by a metric or a norm is the topology for which the the set of open balls is a base. Two metrics (norms) are equivalent iff they induce (generate) the same topology .

Metrics $d,e$ are called uniformly equivalent iff there exists positive $k_1,k_2$ such that $k_1d(x,y)\leq e(x,y)\leq k_2d(x,y).$ Uniformly equivalent metrics are equivalent metrics, but two metrics can be equivalent without being uniformly equivalent.

On the other hand if two norms $\|\cdot \|_1$ and $\|\cdot \|_2$ on a real or complex vector space are equivalent they must be uniformly equivalent, in the sense that the metrics $d_1(x,y)=\|x-y\|_1$ and $ d_2(x,y)=\|x-y\|_2$ are unformly equivalent metrics.

The standard topology on $\mathbb R$ is also called the usual topology on $\mathbb R.$


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