# Construction of an Open, Dense, Connected Set in the Plane

I'm stumped with the following problem.

Let $$\varepsilon>0$$ be given. Prove that there exists an open, dense, and connected set $$G\subset \mathbb{R}^{2}$$ such that $$m_{2}(G)<\varepsilon$$, where $$m_{2}$$ is the Lebesgue measure on $$\mathbb{R}^{2}$$.

My thoughts: I'm thinking that I need to use some sort of construction with a Cantor-like set in $$\mathbb{R}^{2}$$ and then take a set complement. However, I haven't worked with Cantor sets outside of $$\mathbb{R}$$, so I'm not sure what constitutes a "Cantor-like set" in higher dimensions (if this is even defined or a valid construction) Is this roughly what I should want to do? Otherwise, I'm not sure where I should start.

Thanks in advance for any help!

• Draw a sombrero containing $\mathbb{R}$. Then draw a bunch of vertical sombreros chosen well. – T. Bongers Oct 20 '18 at 15:15

Here's a quick outline of a "brute force" construction:

• One way to make a dense open set is to enumerate the points both of whose coordinates are rationals (more generally, in $$\mathbb{R}^n$$ we want to enumerate $$\mathbb{Q}^n$$) as $$(q_i)_{i\in\mathbb{N}}$$, and then put an open ball $$B_i$$ around each $$q_i$$. Since the rationals are dense, the open set $$B=\bigcup B_i$$ will be dense. Now, do you see a way to pick balls so that $$B$$ has "small" measure?

• Now the result of the above won't be connected (exercise). So we need to make it connected. The idea now is to put "bridges" between the open balls we've already drawn - given $$B_i, B_j$$, fix points $$x_i, x_j$$ in each ball (say, their centers) and consider some open set $$L_{i,j}$$ around the line segment connecting $$x_i$$ and $$x_j$$. Do you see how to design these $$L_{i,j}$$s so that the sum of their measures is "small"?

Exercise: The claim is obviously false for $$n=1$$, since the only connected open subsets of $$\mathbb{R}$$ are open intervals, and the only dense open interval is $$(-\infty,\infty)$$. So where does the above argument go wrong if we try to run it in $$\mathbb{R}$$?

A tangential comment (hidden since it contains spoilers):

The construction above can ultimately lead you in the direction of "higher" metric spaces; that is, metric spaces whose "points" are more usually thought of as sets of points. Here's how. In my opinion, the simplest approach to the second bulletpoint above is to look at the set of points whose distance to the given line segment is $$<\epsilon$$ for an appropriate $$\epsilon$$. This kind of "ball around a set" (as opposed to point) is a very useful notion in metric spaces. In fact, we can leave points behind entirely (well, not really) and define a "distance" function on arbitrary sets in a metric space, namely the infimum of the distances between a point in one set and a point in the other set. This isn't a metric in general, but is when we restrict to appropriate sets (exercise: convince yourself that we should restrict attention to the compact sets) and shows up in a number of situations. (Going further afield, it turns out that this isn't the only reasonable metric to put on ("nice") subsets of a metric space, but that's not related to the current problem at all; I just think it's neato.)

(What, tangential comments shouldn't be longer than the actual relevant answer? Nonsense I say!)

Here is an outline. The construction is based on the fact that you can draw an infinite sombrero that only contains finite area under it. To be precise, $$\int_{-\infty}^{\infty} \frac{dx}{1 + x^2} < \infty$$.

• Density is handled by putting vertical versions of this set at carefully chosen points.

• Finite area is handled by scaling.

• Connectedness is handled by putting a horizontal version to connect all the vertical ones.