# The Unapologetic Mathematician

## The Topology of Higher-Dimensional Real Spaces

As we move towards multivariable calculus, we’re going to primarily be concerned with the topological spaces $\mathbb{R}^n$ (for various values of $n$) just as in calculus we were primarily concerned with the topological space $\mathbb{R}$. As a topological space, $\mathbb{R}^n$ is just like the vector space we’ve been discussing, but now we care a lot less about the algebraic structure than we do about the notion of which points are “close to” other points.

And it turns out that $\mathbb{R}^n$ is a metric space, so all of the special things we know about metric spaces can come into play. Indeed, inner products define norms and norms on vector spaces define metrics. We can even write it down explicitly. If we write our vectors $x=\left(x_1,\dots,x_n\right)$ and $y=\left(y_1,\dots,y_n\right)$, then the distance is

\displaystyle\begin{aligned}d(x,y)=\lVert y-x\rVert&=\\\sqrt{\langle y-x,y-x\rangle}&=\\\sqrt{\left(y_1-x_1\right)^2+\dots+\left(y_n-x_n\right)^2}&\end{aligned}

Incidentally, this is the exact same formula we’d get if we started with the metric space $\mathbb{R}$ and built up $\mathbb{R}^n$ as the product of $n$ copies.

One thing I didn’t mention back when I put together products of metric spaces is that we get the same topology as if we’d forgotten the metric and taken the product of topological spaces. This will actually be useful to us, in a way, so I’d like to explain it here.

We define the topology on a metric space by using balls of radius $\delta$ around each point to provide a subbase for the topology. On the other hand, when we have a product space we use preimages of open sets under the canonical projections to provide a subbase. To show that these generate the same topology, what we’ll do is show that the identity map from $\mathbb{R}^n$ as a product space to $\mathbb{R}^n$ as a metric space is a homeomorphism. Since it’s obviously invertible, we just need to show that it’s continuous in both directions. And we can use our subbases to do just that.

What we have to show is that each set in one subbase is open in terms of the other subbase. That is, for each point in the set we should be able to come up with a finite intersection of sets in the other subbase that contains the point, and yet fits inside the set we started with.

Okay, so consider the preimage of an open set $U\subseteq\mathbb{R}$ under the projection $\pi_i:\mathbb{R}^n\rightarrow\mathbb{R}$. That is, the collection of all $x=\left(x_1,\dots,x_n\right)$ with $x_i\in U$. Clearly since $U$ is open in the metric space $\mathbb{R}$ we can pick a radius $\delta$ so that the open interval $\left(x_i-\delta,x_i+\delta\right)$ is contained in $U$. But then the ball of radius $\delta$ in $\mathbb{R}^n$ around the point $x$ contains the point, and is itself contained in $\pi_i^{-1}(U)$, for if $|y_i-x_i|>\delta$ for some other point $y$, then $y$ cannot possibly be within the ball of radius $\delta$ around $x$.

On the other hand, let’s take a ball of radius $\delta$ about a point $x=\left(x_1,\dots,x_n\right)$. We set $\epsilon=\sqrt{\frac{\delta^2}{n}}$ and consider the open intervals $U_i=\left(x_i-\epsilon,x_i+\epsilon\right)$. I say that the intersection of the preimages $\pi_i^{-1}(U_i)$ is contained in the ball. Indeed, if $y=\left(y_1,\dots,y_n\right)$ is in the intersection, the furthest any coordinate can be from the center is $|y_i-x_i|<\epsilon$. Thus we can calculate the total distance

\displaystyle\begin{aligned}\lVert y-x\rVert&=\sqrt{\left(y_1-x_1\right)^2+\dots+\left(y_n-x_n\right)^2}\\&<\sqrt{\epsilon^2+\dots+\epsilon^2}\\&=\sqrt{\frac{\delta^2}{n}+\dots+\frac{\delta^2}{n}}\\&=\sqrt{\delta^2}=\delta\end{aligned}

and so the whole intersection must be within the ball.

This approach is pretty straightforward to generalize to the case of any product of metric spaces, but I’ll leave that as an exercise.

September 15, 2009 - Posted by | Point-Set Topology, Topology