Examples of metric spaces which are not normed linear spaces?

Question

Give an example of a metric space which is not a normed linear space. Justify your example.

Answer 1

Hint: Every metric induced by a norm is unbounded.

Answer 2

It seems the following. Put $X=\{0;1\}$ and define a metric $d$ on the set $X$ as follows: $d(x,y)=1$ if $x\not=y$ and $d(x,x)=0$ for each $x,y\in X$. Then $X$ is not a linear space over $\mathbb R$. :-)

PS. Less trivial are examples of linear metrizable spaces admitting no consistent norm.

Answer 3

Hint: Consider metric $$ d:\mathbb{R}\times\mathbb{R}\to\mathbb{R}_+,(x,y)\mapsto |e^x-e^y| $$ Justfy this exmaple yourself.

Answer 4

So first of all formally not a single metric space is a normed space and vice versa: not a single normed space is a metric space. A metric space is a pair $(X,d)$, where $X$ is a set, $d:X\times X\to\mathbb{R}$ is a metric and some conditions are satisfied. While on the other hand a normed space is a pair $(V,\lVert\cdot\rVert)$ where $V$ is a linear space over $\mathbb{R}$ and $\lVert\cdot\rVert:V\to\mathbb{R}$ is a norm that satisfies some conditions.

So as you can see formally one cannot ever be the other. These are simply different mathematical structures, they have different definitions.

That being said, if $(V,\lVert\cdot\rVert)$ is a normed space then there is a standard way of turning it into a metric space $(V,d)$ via $d(v,w):=\lVert v-w\rVert$. And so the way I understand your question is: give an example of a metric space that cannot be turned into a normed space such that the induced metric and the original one coincide.

But this question has a lots of rather simple examples. A normed space $V$ is a linear space over $\mathbb{R}$ and so the cardinality of $V$ is either $1$ or infinite uncountable. In particular no metric space with at least two elements and at most infinite countable cardinality can ever be a normed space. Like for example standard rationals $\mathbb{Q}$.

And what about infinite uncountable? There's an interesting question that given $V$ a linear $\mathbb{R}$-space with a metric $d$ what conditions $d$ has to satisfy so that there is a norm $\lVert\cdot\rVert$ on $V$ which induces $d$. The conditition is well known: that happens if and only if $d$ is translation invariant (meaning $d(v+x,w+x)=d(v,w)$) and absolutely homogenous (meaning $d(rv,rw)=|r|\cdot d(v,w)$). It is not hard to give an example of a metric that doesn't satisfy this (regardless of the linear structure), for example the last condition implies that if $V$ has at least two elements then the metric is unbounded.

Answer 5

Let $\mathbb{R}$ be the space of real numbers and consider the metric given by the following formula $$d(x,y)=\frac{|x-y|}{1+|x-y|},$$ where $x,y\in\mathbb{R}$.

Then $(\mathbb{R},d)$ is a metric space that is not normed.

Let us show that $d(x,y)$ defined above is, indeed, a metric. It is easy to see that

i. $d(x,y)\geq 0, \forall x,y\in\mathbb{R}$,

ii. $d(x,y)=0\Leftrightarrow x=y$,

iii. $d(x,y)=d(y,x), \forall x,y\in\mathbb{R}$.

Therefore, we only have to prove the triangle inequality $$ d(x,z)\leq d(x,y)+d(y,z), \forall x,y,z\in\mathbb{R}.$$ We begin by noting that $$ |x-z|\leq |x-y|+|y-z|,\forall x,y,z\in\mathbb{R}.$$ Then we add $1$ $$ 1+|x-z|\leq 1+|x-y|+|y-z|,\forall x,y,z\in\mathbb{R}.$$ Then, we reverse the terms and multiply by $-1$ to obtain $$ \frac{-1}{1+|x-z|}\leq \frac{-1}{1+|x-y|+|y-z|} $$ which is equivalent to $$ \frac{|x-z|}{1+|x-z|}\leq \frac{|x-y|+|y-z|}{1+|x-y|+|y-z|}. $$ Taking into account that $$\frac{|x-y|+|y-z|}{1+|x-y|+|y-z|}=\frac{|x-y|}{1+|x-y|+|y-z|}+\frac{|y-z|}{1+|x-y|+|y-z|}$$ and that each of the terms on the right side has an upper bound $$ \frac{|x-y|}{1+|x-y|+|y-z|}\leq\frac{|x-y|}{1+|x-y|}$$ and $$ \frac{|y-z|}{1+|x-y|+|y-z|}\leq\frac{|y-z|}{1+|y-z|}$$ we obtain that $$ \frac{|x-z|}{1+|x-z|}\leq \frac{|x-y|}{1+|x-y|}+\frac{|y-z|}{1+|y-z|}, $$ which is equivalent to $$d(x,z)\leq d(x,y)+d(y,z),\forall x,y,z\in\mathbb{R}.$$

However $d(x,y)$ does not come from a norm (or is not norm-induced), because it does not satisfy $$ \left\Vert \lambda x\right\Vert = |\lambda|\left\Vert x\right\Vert, \forall \lambda,x\in\mathbb{R}.$$ Indeed, if we consider the norm defined by $$\left\Vert x\right\Vert=\frac{|x|}{1+|x|},$$ then $$\left\Vert \lambda x\right\Vert = \frac{|\lambda x|}{1+|\lambda x|}=|\lambda|\frac{|x|}{1+|\lambda x|},$$ which is different then $$ |\lambda|\left\Vert x\right\Vert = |\lambda|\frac{|x|}{1+|x|}.$$

In conclusion, $\left\Vert\cdot\right\Vert$ is not an actual norm by definition and so, it can not induce the metric $d$. Thus, we have reached our conclusion.

A useful link is also this one: Metric Proof