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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.

Don Larynx
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    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
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    Not being closed does not imply being open. – Tobias Kildetoft Aug 24 '13 at 14:52
  • @ZevChonoles my book says that metric spaces are sets equipped with real-valued functions for all elements. – Don Larynx Aug 24 '13 at 14:54
  • @TobiasKildetoft what could it imply? Some pictures would help here – Don Larynx Aug 24 '13 at 14:54
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    @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
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    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
  • @ZevChonoles how should I go about solving this exercise? Suggestions? – Don Larynx Aug 24 '13 at 15:00
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    The second question follows from the first. – dfeuer Aug 24 '13 at 15:08
  • How might I prove the complement of a point is open if I can't use intervals? – Don Larynx Aug 24 '13 at 15:10
  • a] http://math.stackexchange.com/questions/203728/every-single-point-is-closed , b] Given any set X, consider all open sets in the metric space that contain X (call them $G_\alpha, \alpha \in A$ where A is some index set) and let $Y = \cap_{\alpha \in A} G_\alpha$. Show that $Y = X$. – TenaliRaman Aug 24 '13 at 15:10
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    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
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    Also, you need to copy down the definition of a metric space and the definition of an open set in a metric space and stare at those from time to time as you think. – dfeuer Aug 24 '13 at 15:16
  • Concerning proof by contradiction: your textbook is absolutely right. Some people do look down on its use when it’s not necessary, but in my view they’re overly picky. Yes, it’s good to be able to recognize when you’ve used it unnecessarily — that’s an indication that you really understand what you’re doing — but a valid proof is a valid proof, and if a (valid!) proof by contradiction is the one that you find, then for you that proof may be the most understandable. – Brian M. Scott Aug 24 '13 at 18:41
  • Just want to say thanks for the answers guys, I'm still comprehending them. – Don Larynx Aug 24 '13 at 19:39
  • To my experience, most reductio ad absurdum arguments used in relatively simple math is simply the contrapositive in disguise. People sometimes seem to prefer this since it is either more convincing (to the unexperienced) or more known than $(p\to q)\iff (\neg q\to \neg p)$. – Pedro Aug 24 '13 at 22:35
  • @PeterTamaroff: the only substantial difference, in classical logic, is the wording. A "contrapositive" type argument is likely to be written in a more informal style, without direct reference to the principle involved: "Theorem: $a\implies b$. Proof: Let $\neg b$ then ... $\neg a$ QED." – dfeuer Aug 24 '13 at 23:04
  • @dfeuer Yes, that was my point. One can usually see "...but this is a contradiction, so..." when they are simply proving the contrapositive. I probably did it a few times, now I avoid it! – Pedro Aug 24 '13 at 23:05
  • I don't see any problem with using a PBC-structured argument to prove a contrapositive. I have seen direct proofs dressed up as PBC because someone started out with the notion of doing PBC and found themselves doing something else without realizing it—that's where I start cutting out great gobs of unnecessary verbiage. – dfeuer Aug 24 '13 at 23:19

3 Answers3

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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 …

dfeuer
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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 $A$.

Neal
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    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
  • I actually find the shrink-wrap approach more intuitive than your suggestion, hence why I thought of it first, but appealing to part (a) is good pedagogy. – Neal Aug 24 '13 at 15:38
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    @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
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    @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
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    This is great @Neal! It helped me gain additional insight into the problem. – Don Larynx Aug 24 '13 at 21:42
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    Considering the intersection of the complements of the points not in $A$ is of particular elegance given de Morgan's laws. –  Aug 29 '13 at 21:29
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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.

Don Larynx
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