Discussion 1.9.b

What about the square metric or discrete metric on $$\mathbb{R}^2$$? Is there any metric on $$\mathbb{R}^2$$ which makes $$\{(x,y): x^2+y^2 < 1\}$$ nonopen?

Olomana
1. The square metric is always less than or equal to the standard metric, so we can use the proof that the set is open under the standard metric (if the standard metric is less than epsilon, then the square metric is also less than epsilon).

2. The discrete metric is always 0 or 1. For epsilon less than 1, there are no points closer than epsilon other than the point itself, so the set meets the definition of "open" under the discrete metric.

Sorry for the verbal arguments, but my patience with LaTex is limited.

Olomana 22:32, March 2, 2011 (UTC)

'''Agreed. Indeed, we found that every set is open with the discrete metric by your second argument, and taking your first argument further we can reason that the square and Euclidean metrics are compatible (they generate the same open sets). That last question is the tougher one, though. --Steven.clontz 23:13, March 2, 2011 (UTC)'''

Olomana (2)
I think I see how to construct a metric under which the set is non-open. We define a function F which maps half of the x-y plane (points not in the set) onto the y-z plane. Something like:

$$F(x,y) = \left\{\begin{array}{l@{:}l}(x,y,0) & x \leq 1 \\ (0, y, x - 1) & x > 1\end{array}\right.$$

The idea is to have points not in the set arbitrarily close to the x-y plane. We define our metric on the plane to be equal to the standard metric on the mapping in 3-space. We haven't shown yet that Euclidean distance in 3-space is a metric, but if we had, this would give us the needed properties for our constructed metric in the plane.

Now we have points in the set (those on the y-axis) for which any epsilon-neighborhood around that point contains points not in the set (non-zero z values in the mapping to the y-z plane). This means that the set is non-open under this metric.

Edit: thanks for the help with the LaTex. My original function was not one-to-one, so it didn't work. However, all we need is one counter-example to show that the set is non-open. We don't need to show non-openness at every point.

Finally, any 3 points in Euclidean space are co-planar. The triangle inequality holds in that plane, therefore with a suitable substitution of coordinates, it holds in any number of dimensions greater than 2.

Olomana 23:50, March 2, 2011 (UTC)

'''No prob with the LaTeX - it's good practice for me. The distance metric $$d:\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R}$$ is indeed a metric in 3-space (we can skim proving details like those in Discussion questions I think), but I your proposed metric, $$d \circ (F\times F):\mathbb{R}^2\times \mathbb{R}^2 \to \mathbb{R}$$ doesn't satisfy (1). Namely, (0,0) and (1,0) have zero distance in your proposed metric. Wait wait, my bad. Let me look at it further. I think it fails the triangle inequality. --Steven.clontz 15:28, March 3, 2011 (UTC)'''

I've mapped $$\mathbb{R}^2$$ onto a subset of $$\mathbb{R}^3$$. Since the triangle inequality holds for any 3 points in $$\mathbb{R}^3$$, it holds in the subset. I'm defining my metric on $$\mathbb{R}^2$$ as the standard $$\mathbb{R}^3$$ metric on the mapping. Therefore the triangle inequality holds for my (admittedly contrived) metric on $$\mathbb{R}^2$$.

Olomana 17:29, March 3, 2011 (UTC)

'''Ah, okay! Now I see it. Very clever. What you did was something I'd never thought of before: your argument says that, in general, if we can embed any set into a metric space (give an injection from your set to the metric space), it induces a natural metric for the original set.'''

'''The visual I get for your function is that the section of the xy plane for $$x>1$$ is chopped off and positioned normal to the xy along the y-axis. So yes, any open ball in 3D space around points on the y-axis would include points corresponding to points where $$x>1$$. Very nice. --Steven.clontz 20:16, March 3, 2011 (UTC)'''

Steven Clontz
I was thinking more about Olomana's second observation and came up with a similar example. Let's embed the plane $$\mathbb{R}^2$$ with standard metric $$d$$ into $$\mathbb{R}^3$$ using the injection

$$f(x,y) = \left\{\begin{array}{l@{\ :\ }l}(x,y,0) & x \text{ is rational} \\ (0, y, \frac{1}{x}) & x \text{ is irrational} \end{array}\right.$$

Let $$D^*$$ be the standard metric for $$\mathbb{R}^3$$, which with $$F$$ induces the metric $$D$$ on $$\mathbb{R}^2$$. Consider any bounded set $$A$$ ($$A$$ is a subset of a standard metric ball of finite radius) which intersects the y-axis. Any $$D$$-ball around a point on the $$y$$-axis includes points whose $$x$$-value is irrational and arbitrarily large, so it cannot be contained in the bounded set $$A$$, and $$A$$ is not open.

What I don't know: can bounded sets which don't intersect the y-axis be open in this metric?

'''Yes, if $$x$$ is rational then $$B_D((x,y), r)$$ where $$r \le |x|$$ cannot contain any irrationals, so it must simply be $$B((x,y), r) \cap \mathbb{Q}$$ in the standard metric. --Ian mi 23:15, March 3, 2011 (UTC)Ian_mi'''

Ah, thanks, you're right. I'd like to see a metric in which every bounded set is not open. (There's always a closed bounded set in a metric space - any singleton.) --Steven.clontz 17:21, March 8, 2011 (UTC)

'''Well, an open ball is always open and bounded in any metric space. This leaves one other questions: can every bounded set in one metric space be not open another? The answer, however, is no since the empty set is open and bounded in every metric space. Ian mi 11:32, March 9, 2011 (UTC)Ian_mi'''

Well, we should remove the trivial example of the empty set. Thinking ahead to topology, we couldn't define a topology on $$\mathbb{R}^n$$ in which the open sets are exactly the unbounded sets (or the empty set), since the intersection of two unbounded sets may be bounded. The indiscrete topology has a single nonempty open set (the whole space) which is unbounded, but it's not metrizable. --Steven.clontz 21:44, March 9, 2011 (UTC)

Olomana (3)
Continuing along these lines, we don't have to use the entire y-axis for our constructed metric. In particular, we could map to any subset of the open interval between y = -1 and y = 1. Since the cardinality of the power set is greater than the cardinality of the set, this means that the cardinality of the number of different metrics on $$\mathbb{R}^2$$ is greater than the cardinality of $$\mathbb{R}^2$$.

Olomana 21:13, March 3, 2011 (UTC)

'''I don't quite see how your cardinality argument (which I agree to) impacts the question at hand. I also don't see what you meant by using the whole y-axis? --Steven.clontz 17:17, March 8, 2011 (UTC)'''

The cardinality isn't really relevant, but it was a surprise to me. The distance metric has properties that looked fairly restrictive, and we had only 3 examples of metrics in the plane. But it turns out that we can construct as many different metrics as we want. There's nothing difficult about satisfying the metric properties.

On the second point, my original construction mapped a half-plane from the x-y plane to the y-z plane, intersecting the x-y plane along the y-axis. I then used the standard metric in 3-space to induce a (contrived) metric on the plane. Nothing in that argument depended on using the whole half-plane or the whole y-axis, and in fact I could have mapped subsets of the half-plane, intersecting in subsets of the y-axis.

Olomana 19:55, March 8, 2011 (UTC)

'''Gotcha. Because there are $$\aleph_2$$-many subsets of $$\mathbb{R}$$, for each $$A \subseteq \mathbb{R}$$, we may define a surjection'''

$$F_A(x,y) = \left\{\begin{array}{l@{\ :\ }l} (0, y, x - 1) & x > 1 \text{ and } y \in A \\ (x,y,0) & \textrm{ otherwise} \end{array}\right.$$

'''which results in $$\aleph_2$$-many metrics on $$\mathbb{R}^2$$. Cool! --Steven.clontz 22:34, March 8, 2011 (UTC)'''