When we say some map $\phi=(\phi_1,\ldots,\phi_n)$ is a continuous map $\mathbb{R}^n\rightarrow\mathbb{R}^n$ we really mean that each component $\phi_i$ is continuous as a function $\mathbb{R}^n\rightarrow\mathbb{R}$ or do we mean something else? $\mathbb{R}^n$ is as a vector space, so we can also consider the distance between any two vectors $\|v-u\|$ and use the definition of continuity on $\mathbb{R}^n$ as a whole. Which approach is the "proper one"? I'm kind of confused.


I know now that there is theorem that says that component-wise continuity is equivalent to continuity. I will try to prove this assertion.

Let $\phi=(\phi_1,\ldots,\phi_n):\mathbb{R}^n\rightarrow\mathbb{R}^n.$

1) Let each of $\phi_i$ be continuous functions from $\mathbb{R}^n$ to $\mathbb{R}$. Let $\|\cdot \|$ denote euclidean norm in $\mathbb{R}^n$. From continuity we have $$\forall i \quad \forall x \quad\forall\epsilon>0\quad \exists \delta >0\quad\forall y :\|x-y\|<\delta\implies|\phi_i(x)-\phi_i(y)|<\epsilon/\sqrt{n}$$

  • 11
    $\begingroup$ It's a good exercise to show the component-wise continuity condition you articulate agrees with the general definition of continuity using the product topology on the target. This is a good sign that the "product topology" is the correct definition for these constructions. $\endgroup$ – hardmath Jul 17 '15 at 14:16
  • $\begingroup$ That is what i thought, I will try to show it $\endgroup$ – luka5z Jul 17 '15 at 14:16
  • $\begingroup$ @hardmath - what if we choose some very abstract norm on $\mathbb{R}^n$? This property is norm-invariant? $\endgroup$ – luka5z Jul 17 '15 at 14:38
  • 2
    $\begingroup$ math.stackexchange.com/questions/57686/… gives you a bit in the direction of understanding that differing norms on a finite dimensional topological space (like $\mathbb{R}^n$) give equivalent topologies and hence limits. Also, see page 67-68 of supermath.info/AdvancedCalculus13.pdf $\endgroup$ – James S. Cook Jul 17 '15 at 14:42
  • 1
    $\begingroup$ If the codomain of $\phi:X\rightarrow\mathbb R^n$ is equipped with the producttopology then $\phi$ is continuous if and only its components (actually compositions $\pi_i\circ\phi:X\rightarrow\mathbb R$ where the $\pi_i:\mathbb R^n\rightarrow R$ denote the projections) are continuous. The producttopology is "initialized" by the projections. It is the coarsest such that the projections are continuous. Fortunately this producttopology coincides with the topology induced by the Euclidean norm. $\endgroup$ – drhab Jul 17 '15 at 14:44

This is one of those cases where category theory clears up the mystery about products of sets with structure. Once the product in $\text Set$ of a collection of sets $\left \{ X_{\alpha } \right \}_{\alpha \in I}$ is obtained, using the universal mapping property (UMP), then if the individual $X_{\alpha }$ have structure---for example, if they have topologies or if they are groups-- then one can say exactly what the product $\mathit must$ be, in order for the UMP to be preserved.

Now, the fact, for example, that $\mathbb R^{\omega }$ is metrizable in the product topology but not in the box toplogy, tells us in some sense, that the product topology is the "right one" to use and the box topology is not if we want nice results; i.e. the UMP is what determines how we should view products. Of course, this does not mean that we can always get what we'd like: for example, if $I$ is uncountable, $\mathbb R^{I}$ is not metrizable in the product toplogy.

As for norms, they are all equivalent on $R^{n}$ for $n\in \mathbb N$, but not on infinite-dimensional spaces. For example, the $l_{p}$ spaces are all different with $l_{p}\subset l_{q}$ whenever $q>p$, the obvious example being the sequence $\left \{ \frac{1}{n} \right \}_{n\in N}$ which is in $l_{2}$ but not in $l_{1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.