A set in $\mathbb{R}^n$ that is open, closed, and non-empty is $\mathbb{R}^n$ Let $S$ be a set in $\mathbb{R}^n$ that is open, closed, and non-empty.  Prove $S = \mathbb{R}^n$.
Proofs of this available.  My question is to verify my proof and respond to the follow up questions at the end.
Proof: Let $s$ be an arbitrary point in $S$.  Let $D = \{x \in \mathbb{R}^+ : B_x(s) \subset S\}$ (with $B_x(s)$ indicating the open ball of distance $x$ around $s$).  We will show $D$ is not bounded.
Since $S$ is open, $D$ is non-empty.  Hence, if $D$ is bounded, $D$ has a supremum, which we call $r$.  We will show that the existence of $r$ leads to a contradiction, and so $D$ cannot be bounded.
(1) For any $d \in D, \overline{B_d(s)} \subset S$, since $S$ is closed.  But, since $S$ is open, this implies there exists an $\varepsilon > 0$ such that $B_{d+\varepsilon}(s) \subset S$ [see footnote*].  Therefore, $r \notin D$, because $r \in D$ would imply there exists $\varepsilon > 0$ such that $(r + \varepsilon) \in D$, a contradiction.
(2) For any $\varepsilon$ such that $0 < \varepsilon < r, (r - \varepsilon) \in D$, by definition of open ball and supremum.  The boundary of $B_r(s)$ therefore consists entirely of limit points of $S$ and is thus in $S$, since $S$ is closed.  As shown above, $\overline{B_r(s)} \subset S$ implies that there exists an $\varepsilon > 0$ such that $B_{r+\varepsilon}(s) \subset S$, again a contradiction.
Thus, $D$ is unbounded, and $S = \mathbb R ^n$.
Questions:

*

*Is my proof correct?

*Is it well-organized and clear? The two cases I examined seemed similar enough that I'd expect I could merge them, but I was unable to do so.

*It uses the axiom of completeness, implying that a metric space that admitted only rational or natural distances could have proper subsets that are open, closed, and non-empty.  Is that true? What's an example?

*At the footnoted*, I implicitly assumed that if every point on the boundary of ball $X$ has a ball around it, their union includes ball $Y$, with $Y$ a proper superset of $X$.  This seems clear geometrically.  Am I allowed to assume it? Are there metric spaces where this is not true, and would they admit sets that are open, closed, non-empty, and proper subsets?

 A: Ok, here's that answer I said I'd write up.

Comments on your answer
This is a solution-verification, so first up : let's look at your answer.

"Let $s$ be an arbitrary point ... and so $D$ cannot be bounded"

May I say that this is the first place that I've seen this argument, and I think it's a really good idea to consider it. Of course, the proof is correct to this point, but it's also novel in that I've not seen it anywhere else.
Cases (1) and (2) are also fine, right until I see the key point underlining each of them, as mentioned in 4., the last part of the post.

If every point on the boundary of a ball $B$ has a ball of some positive radius around it, then the union of all these balls contains a ball which strictly contains $B$ and is concentric with $B$.

I felt that this statement was false, as written. However, it is in fact, correct. The justification, however, is somewhat difficult (at least at your level), which is why I was surprised that the statement was made with such confidence.
Let me try and list why I naively felt this was wrong. It is true that there is a ball of some positive radius around each point : however, if these radii happened to shrink to zero, then we should be in trouble, right?
Here's an example : if $d=2$, and we consider the unit ball $\{(x,y) : x^2+y^2 = 1\}$. This can be parametrized by $\{(\cos \theta, \sin \theta) : 0\leq  \theta < 2\pi\}$. I was thinking along these lines : well, what if we had , like, circles of radius $\theta$ around every point $\theta \neq 0$, and then we had like a circle of radius $\frac 1{100}$ or something at $0$?
Something along those lines : the radii are positive, but their infimum is equal to zero. I thought that was wrong.
Turns out I'm wrong , so I'll put my final thoughts on your attempt this way.

(1) and (2) are correct, and therefore the entire proof is correct. However, the justification for the Footnote* is perhaps too implicit at the level of OP, and should either receive further elaboration or be replaced by a more elementary assertion.

It is quite surprising that your attempt should create as much debate , but I'd just like users to visualize what I was trying to get at. Perhaps they shouldn't try to visualize why I'm wrong, because I think that's a fairly abstract argument.

Here is a proof of the Footnote* :

Lemma : Let $B(x,r) \subset \mathbb R^d$ be a ball. For every point $y$ on the boundary of $B(x,r)$ (rigorously, this is the set $\{y : \|y-x\| = r\}$), let $r_y>0$ be a positive number (the "radius corresponding to $y$"). Let $$S = B \cup \left[\bigcup_{y} B(y,r_y)\right]$$ Then, there is a $R>r$ such that $B(x,R) \subset S$.

Proof : I think the easiest is the one that I suggested all the way at the starting, and uses the Bolzano-Weierstrass theorem : any bounded sequence of elements in $\mathbb R^d$ has a convergent subsequence. It also uses the continuity of Euclidean distance.
Suppose that there is no $R>r$ such that $B(x,R) \subset S$. Then, for every $\frac 1n$, $B(x,r+\frac 1n) \not \subset S$, hence there is an element $x_n \in B(x,r+\frac 1n)$ such that $x_n \in S^c$.
Now, by choice, every $x_n \in B(x,r+1)$. Therefore, the $x_n$ is a bounded sequence, hence has a convergent subsequence, say $x_{n_k} \to x^*$. We claim that $x^*$ is a point on the boundary of $B(x,r)$.
To see this, note that $x_{n_k} \to x^*$, and the continuity of Euclidean distance, implies that $\|x_{n_k} - x\| \to \|x^*-x\|$. However, note that $r< \|x_{n_k}- x\| \leq r+\frac 1{n_k}$. Therefore, as $k \to \infty$, by the squeeze theorem, $\|x_{n_k} - x\| \to r$. Hence, $\|x^*-x\| =r$.
However, we claim that $x^*$ contradicts the hypotheses. Indeed, notice that $B(x^*,r_{x^*})$ must contain at least one of the $x_{n_k}$ because $x_{n_k} \to x^*$. However, $x_{n_k} \notin S$, therefore it is impossible that $B(x^*,r_{x^*}) \subset S$, contradicting the positivity of $r_{x^*}$.
Therefore, there is an $R>r$ such that $B(x,R)\subset S$, completing the proof.

Sam suggested a proof using "straight line" intuition instead of circles. That also makes sense : indeed, spheres in $\mathbb R^d$ grow out in all directions. If you can instead show that every possible line is contained in $S$, then you're done as well. This turns out to be a more geometrically easy proof, for various reasons.
To do that , let $l$ be any line and let $a$ be a point on $l$. Think of the line $l$ as an isometric image of $\mathbb R$ : that is, if we have two points $x,y$ on $l$, let $d(x,y)$ be the distance between $x,y$ , as if they were in $\mathbb R$. Now we will treat points on $l$ like real numbers.
Now, let $r = \sup\{\epsilon > 0 : x \in l , d(x,a) < \epsilon \implies x \in S\}$. It is geometrically obvious that $r = \infty$. Indeed, suppose not and let $r$ be finite. As $S$ is closed, we can find a "boundary point" i.e. a point $x^* \in S \cap l$ such that $d(x^*,a) = r$. However, $x^* \in S$ and $S$ is open, hence there is a ball of radius $B(x^*,r_{x^*}) \subset S$. This ball will intersect the line $l$, and travelling along the line $l$ it is clear that this ball contains points that are

*

*in $S \cap l$, and


*further away from $a$ than $x^*$, by travelling from $x^*$ in the direction away from $a$.
It follows rather obviously that $r$ cannot be the supremum from here, a contradiction. Hence, $r = +\infty$. Repeating this for all lines $l$ containing an initial choice of point $a \in S$ (by assuming $S$ is non-empty) tells you that $S = \mathbb R^d$.

ADDENDUM :
Indeed, the property discussed in the hypothesis : the non-existence of a non-empty, closed and open set $S$ which is not equal to $\mathbb R^d$, is equivalent to a property called the connectedness of $\mathbb R^d$. A space is connected if it doesn't have a non-empty proper subset which is both open and closed.
Connectedness is a purely topological notion, that does not rely upon geometric notions such as distances, balls etc.
Even though the above definition of connectedness is useful, it is typically not as utilizable in the above form as it is in this form(although both are easily equivalent) :

A topological space is connected if and only if there is no continuous map $f$ from it to $\mathbb R$ taking only the values $0$ and $1$, such that both $f^{-1}(\{0\})$ and $f^{-1}(\{1\})$ are non-empty open sets. (This is typically stated in a different way, but I avoid that for convenience)

Using the above , one can use calculus to show that :

$\mathbb R$ is connected.

Proof : If $f$ is any such map, then let $f(x) = 0$ and $f(y)=1$ for some $x,y$. Clearly $x \neq y$. By the intermediate value theorem, $f$ is continuous on the interval between $x$ and $y$, therefore there must be a $z$ between these points such that $f(z) = \frac 12$. This is a contradiction.

A product of two non-empty connected topological spaces is connected.

Let $A,B$ be non-empty connected topological spaces, and let $f : A \times B \to \mathbb R$ be a map satisfying the conditions. Then, let $a \in A$, and let $f_a(b) = f(a,b)$. Then $f_a : B \to \{0,1\}$ is also seen to satisfy the conditions relative to $B$. It follows that $f_a$ is either constantly $0$ or constantly $1$ for every $a$. Similarly, defining $f_b(a) = f(a,b)$ for every $b \in B$, each $f_b$ is constant.
However, if $a \neq a'$, then $f_a(b) = f_b(a) = f_b(a') = f_{a'}(b)$ , so $f_a = f_{a'}$. Thus, $f$ is in fact a constant map.
In this way, we can show that $\mathbb R \times \mathbb R = \mathbb R^2$ is connected. Proceeding upwards by induction, every $\mathbb R^d$ is connected.
Note that this approach is extremely general, and also solves the problem.

NOTE : I haven't been able to make the compactness argument rigorous yet. I am not planning to do it because it seems infeasible in higher dimensions to be explicit about estimates and I don't want to handwave such an argument either.
A: Great discussion! Very appreciative of such a vigorous response.
Responding to points 3 and 4 in the OP:



*...implying that a metric space that admitted only rational or natural distances could have proper subsets that are open, closed, and non-empty. Is that true? What's an example?





*...Are there metric spaces where [it] is not true [that if every point on the boundary of ball X has a ball around it, their union includes ball Y, with Y a proper superset of X], and would they admit sets that are open, closed, non-empty, and proper subsets?


Wikipedia lists two examples of non-trivial clopen sets, one which corresponds to point 3 and one corresponding to point 4:


*The set of all positive numbers whose square is greater than 2 is clopen in $\mathbb{Q}$ but not in $\mathbb{R}$ for exactly the reason anticipated in point 3.


*In the metric space within $\mathbb{R}$ of $(0,1) \cup (2,3)$, the set $(0,1)$ is clopen.  I believe that this is because this space does admit such a thing.  For example, consider $b = B_{\varepsilon}(3)$. The union of balls around each point on $b$'s boundary is not a proper superset of $b$.
This suggests that A complete metric space with non-trivial clopen sets must not have the union-of-balls-around-balls-is-a-proper-superset property.  Is that true? Does the proof (with @SarveshRavichandranIyer 's additions) prove it?
