Norms on Rd
Definition (Family of \(p\) norms on \(\mathbb{R}^{d}\))
On \(\mathbb{R}^{d}\) we consider the following norms:
- Euclidean: \(\|(x_{1},\dots,x_{d})\| = (x_{1}^{2}+\dots+x_{d}^{2})^{1/2}\).
-
Supremum norm: $$|(x_{1},\dots,x_{d})|_{\infty} = \max{ x_{i} : 1 \leq i \leq d}$$. -
\(p\) norm (\(1 \leq p < \infty\)): $$|(x_{1},\dots,x_{d})|_{p} = ( x_{1} ^{p}+\dots+ x_{d} ^{p})^{1/p}$$.
The case \(p = 2\) coincides with the Euclidean norm, that is,
\[\|(x_{1},\dots,x_{d})\| = \|(x_{1},\dots,x_{d})\|_{2}.\]Definition (Ball associated with a \(p\) norm)
For \(x_{0} \in \mathbb{R}^{d}\) and \(r > 0\) we define
\[B_{p}(x_{0},r) = \{ x \in \mathbb{R}^{d} : \|x - x_{0}\|_{p} < r \}.\]When \(p = 2\) we simply write \(B(x_{0},r)\).
Definition (Open set in \(\mathbb{R}^{d}\))
\(D \subseteq \mathbb{R}^{d}\) is an open set if for every \(x_{0} \in D\) there exists \(r > 0\) such that
\[B(x_{0},r) \subseteq D.\]In particular \(\emptyset\) and \(\mathbb{R}^{d}\) are open, and every ball is open.
Equivalence of p norms on Rd
Proposition (Bounds between the Euclidean norm and the supremum norm)
For every \((x_{1},\dots,x_{d}) \in \mathbb{R}^{d}\),
\[\|(x_{1},\dots,x_{d})\|_{\infty} \leq \|(x_{1},\dots,x_{d})\| \leq \sqrt{d}\,\|(x_{1},\dots,x_{d})\|_{\infty}.\]| Proof: For \(1 \leq i \leq d\), $$ | x_{i} | \leq \sqrt{x_{1}^{2}+\dots+x_{d}^{2}} = |(x_{1},\dots,x_{d})|\(, which gives\)|(x_{1},\dots,x_{d})|{\infty} \leq |(x{1},\dots,x_{d})|$$. Conversely, |
Proposition (Bounds between the \(p\) norm and the supremum norm)
For every \(1 \leq p < \infty\) and every \((x_{1},\dots,x_{d}) \in \mathbb{R}^{d}\),
\[\|(x_{1},\dots,x_{d})\|_{\infty} \leq \|(x_{1},\dots,x_{d})\|_{p} \leq d^{1/p} \|(x_{1},\dots,x_{d})\|_{\infty}.\]Consequently \(\|(x_{1},\dots,x_{d})\| \leq \sqrt{d}\,\|(x_{1},\dots,x_{d})\|_{p}\).
| Proof: For each \(1\leq i\leq d\), $$ | x_{i} | ^{p}\leq \sum_{j=1}^{d} | x_{j} | ^{p}\(, and taking\)p$$-th roots yields the lower bound. For the upper bound, |
and taking \(p\)-th roots concludes the argument. The last inequality combines both with the Euclidean bound \(\leq \sqrt{d} \|\cdot\|_{\infty}\).
Lemma (The \(p\) norms define the same open sets in \(\mathbb{R}^{d}\))
The norms \(\|\cdot\|_{p}\) (\(1\leq p\leq\infty\)) determine the same family of open sets in \(\mathbb{R}^{d}\).
Proof: It suffices to show that given \(r > 0\) there exists \(r_{p} > 0\) with \(B_{p}(x,r_{p}) \subseteq B(x,r)\) and, conversely, there exists \(r' > 0\) with \(B(x,r') \subseteq B_{p}(x,r)\). By the previous proposition, if \(\|x-y\|_{p} < r/\sqrt{d}\) then \(\|x-y\| \leq \sqrt{d}\,\|x-y\|_{p} < r\), so that \(r_{p} = r/\sqrt{d}\) satisfies
\[B_{p}\!\left(x,\tfrac{r}{\sqrt{d}}\right) \subseteq B(x,r).\]Analogously, \(B(x, r/d^{1/p}) \subseteq B_{p}(x,r)\). Hence every set open with respect to one norm is open with respect to the other.
Connectedness and path-connectedness in Rd
Definition (Disconnected and connected set)
\(G \subseteq \mathbb{R}^{d}\) is disconnected if there exist open sets \(G_{0}, G_{1}\) such that
\[G_{0} \cap G \neq \emptyset, \quad G_{1} \cap G \neq \emptyset, \quad G_{0} \cap G_{1} = \emptyset \quad\text{and}\quad G \subseteq G_{0} \cup G_{1}.\]A set is said to be connected if it is not disconnected.
Definition (Curve)
A curve is a continuous function \(\gamma : [a,b] \to \mathbb{R}^{d}\).
Definition (Path-connected set)
\(E \subseteq \mathbb{R}^{d}\) is path-connected if for all \(x_{0}, x_{1} \in E\) there exists a curve \(\gamma : [a,b] \to \mathbb{R}^{d}\) such that
\[\gamma(a) = x_{0}, \quad \gamma(b) = x_{1} \quad\text{and}\quad \gamma(t) \in E \text{ for every } t \in [a,b].\]Note (Reparametrisation to the interval \([0,1]\))
If \(\gamma : [a,b] \to \mathbb{R}^{d}\) is continuous, then \(\gamma_{1} : [0,1] \to \mathbb{R}^{d}\) given by \(\gamma_{1}(s) = \gamma((b-a)s + a)\) is continuous. Hence, in the definition of path-connectedness one may assume \(a = 0\) and \(b = 1\).
Lemma (A path-connected set admits no decomposition into disjoint open sets)
Let \(E \subseteq \mathbb{R}^{d}\) be path-connected. Then there exist no non-empty open sets \(G_{0}, G_{1}\) such that \(E \subseteq G_{0} \cup G_{1}\) and \(G_{0} \cap G_{1} = \emptyset\).
| Proof: Suppose such \(G_{0}, G_{1}\) exist and take \(x_{0} \in G_{0} \cap E\), \(x_{1} \in G_{1} \cap E\). By path-connectedness there exists \(\gamma : [0,1] \to E\) with \(\gamma(0)=x_{0}\) and \(\gamma(1)=x_{1}\). Since \(G_{0}\) is open, there exists \(r > 0\) with \(B(x_{0}, r) \subseteq G_{0}\). By continuity of \(\gamma\) there exists \(\delta > 0\) such that $$ | t | < \delta\(implies\)|\gamma(0) - \gamma(t)| < r\(, so that\)\gamma(t) \in B(x_{0}, r) \subseteq G_{0}\(for\)0 < t < \delta$$. |
Define
\[t_{0} = \sup \{ t > 0 : \gamma(s) \in G_{0},\ 0 \leq s < t \}.\]| The set is bounded by \(1\) and contains \(\delta/2\) (since \(\gamma(s) \in G_{0}\) for \(0 \leq s < \delta\) by the above), so \(\delta/2 \leq t_{0} \leq 1\). If \(\gamma(t_{0}) \in G_{0}\), there exists \(r_{1} > 0\) with \(B(\gamma(t_{0}), r_{1}) \subseteq G_{0}\) and, by continuity, there exists \(\delta_{1} > 0\) with $$ | t_{0} - s | < \delta_{1} \implies \gamma(s) \in B(\gamma(t_{0}), r_{1}) \subseteq G_{0}$$. This contradicts the definition of the supremum. |
| Hence \(\gamma(t_{0}) \in E \setminus G_{0} \subseteq G_{1}\). But then there would exist \(r_{2}, \delta_{2} > 0\) such that $$ | t_{0} - s | < \delta_{2} \implies \gamma(s) \in B(\gamma(t_{0}), r_{2}) \subseteq G_{1}\(. By definition of the supremum there exists\)s’ \in (t_{0} - \delta_{2}, t_{0})\(with\)\gamma(s’) \in G_{0}\(; but then\)\gamma(s’) \in G_{0} \cap G_{1} = \emptyset$$, a contradiction. |
Theorem (Relation between connectedness and path-connectedness in \(\mathbb{R}^{d}\))
Let \(G \subseteq \mathbb{R}^{d}\). Then:
- If \(G\) is path-connected, then \(G\) is connected.
- If \(G\) is open and connected, then \(G\) is path-connected.
Proof: Part (1) is a direct consequence of the previous lemma. Part (2) is proved by fixing \(x_{0} \in G\) and considering the set \(A = \{ x \in G : x \text{ can be joined to } x_{0} \text{ by a curve in } G\}\); using that \(G\) is open one shows that \(A\) and \(G \setminus A\) are both open, and by connectedness together with \(A \neq \emptyset\) one concludes \(A = G\).