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Let $L$ be the language $L=\{<,=,+,-,\cdot, 0,1\}$, with standard interpretations, and let $\mathcal{A}=\langle\mathbb{R}, <,=,+,-,\cdot,0,1\rangle$. Let $S\subseteq\mathbb{R}^n$. Show that if $S$ is definable, then the topological closure of $S$, given as $$\bar{S}:=\{a\in\mathbb{R}^n:\text{every open ball centered at $a$, contains a point of $S$}\},$$ is also definable.

My attempt at a solution:

Clearly, the set $\bar{S}$ is the set of all points in $S$ as well as the boundary of $S$. So really, what we want is the union between the set $S$, and the set $S'$ which I use to denote the set of all boundary points of $S$. We already have that $S$ is definable by some $L$-formula in the given structure, say $\phi^{\mathcal{A}}$. It remains to show that the boundary $S'$ is also definable.

The boundary is the set of all points $a\in\mathbb{R}^n$, so that the following holds:

  • $a\notin S$
  • $\forall\epsilon>0\exists s\in S$ such that $d(a,s)<\epsilon$, where $d$ is the distance function between two points

Let's call the set of elements satisfying the latter property $B$. Clearly the set we are seeking is, $$\phi^{\mathcal{A}}\cup(\mathbb{R}^n\setminus\phi^{\mathcal{A}}\cap B).$$ It remains to show that $B$ is definable then. This is where I'm having some trouble. I can't see how to express this in my given language.

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    $\begingroup$ What difficulty do you have defining $B$? Which part of it do you not know how to express? $\endgroup$ – Eric Wofsey Dec 5 '18 at 20:54
  • $\begingroup$ I think I figured it out. I think the $\psi$ which defines $B$ is given by $$\psi:=\exists x_1...\exists x_n(\phi\wedge\forall\epsilon((0<\epsilon)\to((x_1-y_1)\cdot(x_1-y_1)+...+(x_n-y_n)\cdot(x_n-y_n)<\epsilon\cdot\epsilon))).$$ Still, I'm not too sure that is works. Basically I'm trying to say that there is some $s\in S$ (i.e. satisfying $\phi$), and that in addition to this, the distance between the two is less than epsilon for all epsilon. $\endgroup$ – quanticbolt Dec 5 '18 at 21:04
  • $\begingroup$ You are on the right lines in your comment using the dot product to express the metric, but your quantification is wrong: informally, $\mathbf{y} \in S \cup S'$ iff for any $\epsilon > 0$, there is a $\mathbb{x} \in S$, with $d(\mathbb{x}, \mathbb{y}) < \epsilon$. Now translate that into logical notation (the $\forall$ should come outside the $\exists$, not inside). $\endgroup$ – Rob Arthan Dec 5 '18 at 21:12
  • $\begingroup$ So can I say: $$\forall\epsilon((0<\epsilon)\wedge\exists x_1...\exists x_n(\phi\wedge(x_1-y_1)\cdot(x_1-y_1)+...+(x_n-y_n)\cdot(x_n-y_n)<(\epsilon\cdot\epsilon)))$$ $\endgroup$ – quanticbolt Dec 5 '18 at 21:15
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    $\begingroup$ @quanticbolt What if $\epsilon = 0$? $\endgroup$ – Alex Kruckman Dec 6 '18 at 2:18
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As you wrote in the question, the closure $\overline{S}$ is definable by $(\forall \varepsilon > 0) (\exists y\in S)\, d(x,y)<\varepsilon$. Expanding this into a first-order formula, given that $S$ is definable by $\varphi(x_1,\dots,x_n)$, we have:

$$\forall \varepsilon\, ((\varepsilon > 0) \rightarrow \exists y_1\dots \exists y_n\, (\varphi(y_1,\dots,y_n)\land ((x_1-y_1)\cdot (x_1-y_1) + \dots + (x_n-y_n)\cdot (x_n-y_n) < \varepsilon \cdot \varepsilon))).$$

You were on the right track in the comments, but you weren't expressing the bounded quantifiers correctly.

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