How is $\mathbb{R}^n$ a compact subset of $\mathcal{H}(\mathbb{R}^n)$ (the set of all compact subsets of $\mathbb{R}^n$)?

Question

I'm reading Essential Real Analysis by Michael Field. There's a definition of a metric space $(\mathcal{H}(\mathbb{R}^n),h)$, in which the set $\mathcal{H}(\mathbb{R}^n)$ is the set of all non-empty compact subsets of $\mathbb{R}^n$. Field writes on page 330:

"... In particular, we can regard $\mathbb{R}^n$ as a subset of $\mathcal{H}(\mathbb{R}^n)$ by the map $(x_1,...,x_n)\rightarrow\{(x_1,...,x_n)\}$".

I have difficulties in understanding this. Does he mean that $\mathbb{R}^n$ can be divided into closed and bounded (thus compact) sets and $\mathbb{R}^n\subset\mathcal{H}(\mathbb{R}^n)$ because those compact sets are all in $\mathcal{H}(\mathbb{R}^n)$? But then what is the role of the map he gives?

(h is the Hausdorff distance. It doesn't seem to have anything to do with the claim, tough.)

Answer 1

Points are compact. This makes allows us to consider every subset of $\Bbb R^n$ a subset of $\mathcal H(\Bbb R^n)$ by identifying the point $(x_1,\ldots,x_n)\in\Bbb R^n$ with the point $\{(x_1,\ldots,x_n)\}\in\mathcal H(\Bbb R^n)$. Note that this map $\Bbb R^n\to \mathcal H(\Bbb R^n)$ is not only injective, but is in fact an isometry! So we obtain $\Bbb R^n$ in a very nice manner as a subspace of $\mathcal H(\Bbb R^n)$.

Answer 2

No, it means that $\Bbb R^n$ can be identified (since points are compact) with the set $\widetilde{\Bbb R^n}\subseteq \mathcal H(\Bbb R^n)$ of the singletons of its elements, and (presumably, reading on) that the map $(\Bbb R^n,d)\to \left(\widetilde{\Bbb R^n},h\right)$ sending $x\mapsto \{x\}$ is an isometry.

Answer 3

Your title (at the time of my answer) begins with "How is $\mathbb{R}^n$ a compact subset of $\mathcal{H}(\mathbb{R}^n)$", which is a bit misstated.

In the text of your question you define a certain function, let's call it $f,$ from $\mathbb{R}^n$ into $\mathcal{H}(\mathbb{R}^n).$ Thus, each point in $\mathbb{R}^n$ is associated with a certain compact subset of $\mathbb{R}^n.$ We can also consider the image of subsets of $\mathbb{R}^n$ under this function via the following: Given $E \subseteq \mathbb{R}^n,$ define $f[E] \subseteq \mathcal{H}(\mathbb{R}^n)$ by $f[E] = \{f(e): e \in E\}.$ This is usually called the image of $E$ under $f$ (also called the image under $f$ of $E).$ Since this is taking place in the context of metric spaces and $f$ can be shown to be continuous (can you prove this?), we know that if $E$ is a compact subset of $\mathbb{R}^n,$ then $f[E]$ will be a compact subset of $\mathcal{H}(\mathbb{R}^n).$

However, you are asking about $f[\mathbb{R}^n],$ which is the image of the non-compact subset $E = \mathbb{R}^n$ under $f,$ so the result I just stated does not apply. Moreover, for continuous functions from one metric space $X$ to another metric space $Y,$ it is possible for the image of a non-compact set to be compact (consider a constant function when $X = Y = {\mathbb R})$ and it is possible for the image of a non-compact set to be non-compact (consider the identity function when $X = Y = {\mathbb R}).$

Thus, a natural question is whether $f[\mathbb{R}^n]$ is a compact subset of $\mathcal{H}(\mathbb{R}^n).$ At this point I'll end with two questions that I think would be instructive to work on.

1. Prove or disprove: $f[\mathbb{R}^n]$ is a non-compact subset of $\mathcal{H}(\mathbb{R}^n).$

2. Prove or disprove: For each $E$ such that $E$ is a non-compact subset of $\mathbb{R}^n,$ we have that $f[E]$ is a non-compact subset of $\mathcal{H}(\mathbb{R}^n).$