# Homeomorphism between a normed space and its open unit ball

I have studied that every normed space $X$ is homeomorphic to its open unit ball $B$. I want to know what conclusion can we draw from this statement? Does it mean that every normed space is open ball? I am confused with this statement. Could anyone clear my doubt?

Thanks for help

• From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont. – dantopa Jan 11 at 6:03

It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:B\to X$ defined by $f(x)=\tan(\frac{\pi}{2}\|x\|)\cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X=\{0\}$. To see this, take any $x\neq 0$. Then $x/2\|x\|\in B$ but $2\cdot x/2\|x\|\notin B$.
• As a normed linear space, $B$ isn't. – Robert Israel Jul 4 '12 at 5:43
• $B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not). – Robert Israel Jul 4 '12 at 17:02