by Candid Moon _Max_
Last Updated July 17, 2019 12:20 PM - source

The interval $[0, 1]$ is commonly called the 'unit interval'. Is there something similar for $[-1, 1]$? Like a pre-defined name.

I don't think so. Personally speaking, I have never found a particular definition for intervals of the form $[a,b]$ in general. I believe that the reason of that is that any closed interval is homeomorphic to $[0,1]$ and then it shares the same topological properties of $[0,1]$.

Any pair of closed intervals $[a,b]$ and $[c,d]$ are homeomorphic, for any choice of $a,b,c,d\in\Bbb R$. In fact the function $f:[a,b]\to [c,d]$ defined as $$f(x)=\frac{x-a}{b-a}(d-c)+c$$ is a possible homeomorphism.

By taking $c=0$ and $d=1$, you find an explicit homeomorphism between $[a,b]$ and $[0,1]$.

Also, the unit interval appears in more contexts than a general closed interval. For instance in algebraic topology an homotopy is parametrised by $[0,1]$. Another example comes from probability theory. The probability in measure by a real number in the unit interval $[0,1]$. So it makes sense that $[0,1]$ has a privileged role and has an own definition.

July 17, 2019 12:00 PM

It's the closed unit ball of $\mathbb{R}$ in its usual absolute value norm.

In any normed space $(X, \|\cdot\|)$, we have a closed unit ball (or disk) $$D=\{x \in X: \|x\| \le 1\}$$ and a unit sphere

$$S=\{x \in X: \|x\|=1\}$$ The unit sphere in the reals is $\{-1,1\}$ of course, the boundary of the unit ball $[-1,1]$.

They often get indexed by their dimension $n$ if it is finite, so $D^1$ and $S^0$ in the above case. $\partial D^n = S^{n-1}$ for all $n\ge 1$.

July 17, 2019 12:13 PM

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