# How can open spheres be shapes like squares?

I'm reading through Schaum's General Topology and I came across the section on equivalent metrics. So, the author introduces the idea that metrics $$d_1$$ and $$d_2$$ are equivalent when the $$d-$$open spheres of $$d_1$$ and $$d_2$$ induce the same topology on a space $$X$$. Then, as an example, the author goes on to show that the three metrics below all serve as a base for the natural topology on $$\mathbb{R}^2$$.

So, I can clearly understand how we can use these shapes as a base for the natural topology on $$\mathbb{R}^2$$. What I don't understand is how $$S_{d_1}(p,\delta)$$ and $$S_{d_2}(p,\delta)$$ are open "spheres". Geometrically speaking $$S_{d_1}(p,\delta)$$ and $$S_{d_2}(p,\delta)$$ are squares. Since they are squares, they are not "spheres" in the sense that they are sets centered at some point $$p$$ which contain all points within $$\delta$$ of $$p$$. Like I said, squares are perfectly fine as a base for the natural topology on $$\mathbb{R}^2$$ but can we really call $$S_{d_1}(p,\delta)$$ and $$S_{d_2}(p,\delta)$$ open spheres? In a square centered at $$p$$, won't some points be farther from $$p$$ than others, thus violating our traditional notion of open sphere?

Spheres in metrics do not mean a physical sphere, i.e. a round circle or something like that.

In fact, usually, a sphere is an open neighborhood of a certain point that contains all points which have a distance to the center point lower than $$\delta$$. The key observation is that the "distance" is just a function (which has certain properties s.t. it's called "metric"), but it doesn't have to be the Euclidean distance.

You are imaging a circle - that is using the Euclidean distance as your metric. However, the square comes by using the metric $$d((x_1,y_1),(x_2,y_2)) = \max\{|y_2-y_1|,|x_2-x_1|\}$$. The shape of the set of all points with $$d(x,y) < 1$$ is indeed a square, which is, nevertheless, called the "sphere" around a point. That may seem confusing at first glance, but working with topology, spheres can have all kind of shapes, which could be non-circular as well.

In general open spheres, topologically speaking, are subsets in a metric space $$(R,d)$$ of the form $$S_d(p,\delta) = \{q \in R| d(p,q)<\delta\}$$, whereas closed spheres are $$S_d(p,\delta) = \{q \in R| d(p,q)\leq\delta\}$$.

In your case we have e.g. $$R=\mathbb{R}^2$$.

• I agree with everything you've said, but looking at the example, if the author is saying that the class of open squares serves as a base for the natural topology on $\mathbb{R}^2$, don't we HAVE to assume that the distance is Euclidean distance? – SunRoad May 15 at 13:00
• Yeah, that seems correct. You can imagine that you just need a somehow "small" surrounding of a point in $\mathbb{R}^2$, how this looks like doesn't really matter. – LegNaiB May 15 at 13:12
• @SunRoad Yes that is completely true, it doesnt even have to be convex. the interior of any polygon will work. You just need for the shape to satisfy two things: It needs to be open, and for every open ball you need to be able to place the shape inside the open ball so that it contains the center. – Yorch May 15 at 13:13
• Oh, I just saw how it could be possible even with non-convex shapes. As you can choose the shapes to be arbitrarily small and you can translate them to left/right and up/down, it should be possible. This is no proof, but intuitively I got the point. I now think it's true as well :) – LegNaiB May 15 at 13:26
• Yes, I proved the following: Let $S$ be an open bounded subset of $\mathbb R^2$. Then the family of sets of the form $T(S)$ where $T$ is an affine transform form a basis for $\mathbb R^2$ with the euclidean topology. – Yorch May 15 at 13:26

The metric $$S_{\delta_2}$$ that gives the diamonds is sometimes called the Manhattan distance.

Consider a city in which the points are only the intersections of two streets. Assume that all the city blocks are perfect squares. Then the "distance" between two corners would be the manhattan distance.

So if you are currently located at corner $$c$$ then for every integer distance the set of corners at that distance will form that "diamond" (of course it won't have all of the points since there is only going to be a finite number of points).

So this is a way in which the "spheres" can come up in real life. But I think they are only "spheres" in the sense that they are the sets of points that are at a fixed distance from another "center" point.

• That's really cool and I agree with how you brought up the Manhattan topology. That works precisely when the only points are the ones located at street intersections. But the author seems to be claiming that the open diamonds are "spheres" in $\mathbb{R}^2$, something which is geometrically impossible. – SunRoad May 15 at 12:57
• Seeing my answer that is possible if you interpret "spheres" not as circles, but as a set of points with a certain "distance", where this "distance" function can be different – LegNaiB May 15 at 12:58
• @SunRoad The open diamonds aren't spheres in $\mathbb R^2$ with the euclidean topology. Although perhaps what the author is saying that these open squares can also serve as a "basis" for the euclidean topology. And in fact this is true, there is a very nice result that tells us that all norms for dimensional vector spaces are equivalent ! – Yorch May 15 at 13:12

It depends how you measure distance.

Let's write $$x = (x_{i})_{i=1}^{n}$$ and $$y = (y_{i})_{i=1}^{n}$$ for points in Cartesian $$n$$-dimensional space. If we define $$d(x, y) = \sum_{i} |x_i - y_i|$$, then "spheres" are the axis-oriented "octahedra" as in the right-hand diagram. This metric is induced by a vector space norm, the $$1$$-norm.

More generally, if $$p \geq 1$$ is real we define the "$$p$$-distance" by $$d(x, y) = \bigl(\sum_{i} |x_{i} - y_{i}|^{p}\bigr)^{1/p} =: \|x - y\|_{p}$$. This function satisfies the axioms of a metric (positve-definiteness, symmetry, and the triangle inequality), so is a mathematically consistent measure of distance.

If instead we define $$d(x, y) = \max|x_i - y_i| =: \|x - y\|_{\infty}$$, then "spheres" are axis-oriented cubes as in the middle diagram. Like the $$p$$-distance, this metric is induced by a vector space norm, often called the uniform norm (because of its relationship to "uniform convergence" if you've come across the term). The "$$\infty$$-norm" notation is reasonable, since $$\|\ \|_{\infty} = \lim\limits_{p \to \infty}\|\ \|_{p}$$.