# Metric spaces and openness

I am asked to prove two things. I would like to know if the proof was elaborate and concise.

I would also like to know if proving reductio ad absurdum is looked down upon. I have heard from my professor say it's not recommended, but my analysis book says it is an invaluable tool.

a) Prove that in a metric space the complement of a point is open.

Suppose that the point $x$ is contained in a closed interval [x]. Now suppose for contradiction the complement of the point is closed, that is, $a<=x$ and $x<= b$. Now, the absurdity is at hand: $[x] \cup [a, b] \leftrightarrow [a, b]$ when they were originally meant to be disjoint. So the complement of the point must be open, that is, $a < x < b$.

b) Prove that any set in a metric space is an intersection of open sets.

No clue here. I would gladly appreciate any clues. I am self-learning this.

• This is an arbitrary metric space - not necessarily $\mathbb{R}$. There is no notion of "interval" or "$\leq$". – Zev Chonoles Aug 24 '13 at 14:51
• Not being closed does not imply being open. – Tobias Kildetoft Aug 24 '13 at 14:52
• @Jossie: Yes, a metric space is a set $X$ equipped with a function $d\colon X\times X\to\mathbb{R}$ satisfying certain properties. But the set $X$ can be any set whatsoever - for instance, the set $X=\{\ddot{\smile},\#,\triangle\}$ equipped with the discrete metric is a metric space, and it is not comprised of real numbers. – Zev Chonoles Aug 24 '13 at 14:57
• There exist subsets of a metric space which are neither open nor closed. For instance the interval $[0,1)$ in $\mathbb{R}$. – Dan Rust Aug 24 '13 at 14:57
• Jossie, you need to think more broadishly. A sufficiently rich mental model for this particular problem is the Euclidean plane, $\Bbb R^2$, which has no intervals, but does have distances that respect the triangle inequality. – dfeuer Aug 24 '13 at 15:15

You can do these exercises without reducing to contradiction. They are intended to ensure you understand the concepts and can apply definitions. I would guess that's perhaps why your professor discouraged use of proof by contradiction.

So here are some hints. Let your metric space be $X$. Note that since the context is a general metric space, we cannot appeal to any properties of $\mathbb{R}$, such as intervals (which require, among other things, an ordering). Instead, all we can work with is the distance function and things we've defined from the distance function, like convergent sequences and distance balls.

For part (a), you have a range of definitions of "open" and "closed" at your disposal. To prove that the complement in $X$ of a singleton $\{x\}$ is open, you might prove that for any $y\in X-\{x\}$, there is a small ball of radius $\epsilon > 0$ such that $x\notin B_\epsilon(y)$. Alternately, you might prove that $\{x\}$ is closed (does it contain every limit point?), and argue that the complement of a closed set is open.

For part (b), we are given an arbitrary set $A$. What sets could we intersect? As dfeuer suggests -- is to consider the complement of every point in the complement of $A$. We know these are open sets. Can we recover $A$ from them?

Again, the point here is to help you become comfortable with working in a more abstract setting. You have certain tools at your disposal, like the distance function and its derivative notions, while you cannot appeal to other, more familiar ideas, like intervals.

NB: In an earlier version of my answer, I had included the idea of "shrinkwrapping" $A$. As defeuer, Stefan H., and Brian M. Scott point out in the comments, that only produces the closure of $A$: any point in the boundary of $A$ will be contained in any $\epsilon$-neighborhood of $A$, hence will be in the intersection of all $\epsilon$-neighborhoods of $A$.

• It's much easier to use part (a) to prove part (b) than to use the shrink-wrap approach. I do seem to remember the shrink-wrap approach being good for something else though, probably relating to compactness. – dfeuer Aug 24 '13 at 15:27
• @dfeuer is right in that the shrink wrap approach does not always work, for example each $\varepsilon$-neighborhood of $\Bbb Q\subset \Bbb R$ is the real line, and so is their intersections. If it works, however, then the set will be a $G_\delta$-set. I think for closed sets it always works since the intersection of all these $\varepsilon$-neighborhoods is just $\overline A$. – Stefan Hamcke Aug 24 '13 at 17:23
• @Stefan: In fact in general it gives you $\operatorname{cl}A$, since it gives you $\{x\in X:d(x,A)=0\}$. – Brian M. Scott Aug 24 '13 at 18:49
• This is great @Neal! It helped me gain additional insight into the problem. – Don Larynx Aug 24 '13 at 21:42
• Considering the intersection of the complements of the points not in $A$ is of particular elegance given de Morgan's laws. – user34768 Aug 29 '13 at 21:29

Let $(X,d)$ be a metric space.

Let $x \in X$.

We wish to show that $X\setminus \{x\}$ is open.

By definition, this means showing that for each $y \in X \setminus \{x\}$ there exists an $\epsilon\in(0,\infty)$ such that ….

Fix a specific $y$. Then the distance between $x$ and $y$ is $d(x,y)$.

Let $\epsilon = \text{something or other to do with }d(x,y)$.

Then …

• Hi, is this because we have defined it to be open? That is, can the complement of a point be closed (if we define it to be so)? – Don Larynx Aug 24 '13 at 21:35
• @Jossie, I'm really not sure what you mean. What definition are you using for an open set in a metric space? – dfeuer Aug 24 '13 at 22:41
• Never mind, I misunderstood. – Don Larynx Aug 24 '13 at 22:56

a) Prove that in a metric space the complement of a point is open.

Suppose there exists a metric space $x\in X$. We wish to show $X - {x}$ is open. Thus, define

$$\forall y \in M : D(x, y) < \epsilon$$

By definition, $\forall y \in M, S_{\epsilon}(x)$ is an open ball. Because open balls are open, it has been shown the complement of a point is open.

b) Prove that any set in a metric space is an intersection of open sets.

Suppose there exists a set $E \subset M, E = E_{1} \cup E_{n}$. Next, suppose $e \in E$. Then $e \in E_{n} \forall n$. Since $E_{n}$ is open we have $B_{r_{i}}(e) \subset E_{n}$. If we take $r = min (r_1, r_2, ... r_n)$, then $B_{r}(e) \subset E$. QED.

• "Suppose there exists a metric space $x\in X$" is not sensible. I assume you mean "Suppose $(X,D)$ is a metric space and $x \in X$." – dfeuer Aug 24 '13 at 22:39
• Also, "Thus, define $\forall y\in M:D(x,y)<\epsilon$" does not seem to have any meaning. – dfeuer Aug 24 '13 at 22:40
• @dfeuer I was defining an open ball. – Don Larynx Aug 24 '13 at 22:57
• It's far better to use plain language correctly than to use logical symbols incorrectly. – dfeuer Aug 24 '13 at 23:25
• Was I missing $\epsilon > 0$? – Don Larynx Aug 24 '13 at 23:28