Why discrete set must be countable?

I want to show a set, which every point of it is an isolated point. Then this set must be countable. How to show it?

I find this from Wikipedia's article Isolated point, but I don't understand:

A set that is made up only of isolated points is called a discrete set (see also discrete space). Any discrete subset S of Euclidean space must be countable, since the isolation of each of its points together with the fact that rationals are dense in the reals means that the points of S may be mapped into a set of points with rational coordinates, of which there are only countably many.

• What part don't you understand? – MJD Jun 13 at 18:26
• See for example math.stackexchange.com/q/961689/42969 – Martin R Jun 13 at 18:27
• I don't understand the isolation of each point + rationals are dense part. – Mariana Jun 13 at 18:28
• Depends on the topology on your space. For example if $X$ is any arbitrary set and define any subset of $X$ as open, then your stamens is false (This topology is also metrizable: $d(x,y)=\mathbb{1}(x\neq y)$). – Oliver Diaz Jun 13 at 18:38
• If the dimension is infinite, for example the set of real sequences, it is not true. Take the subset of integer sequences. – Empy2 Jun 13 at 18:52

1. Because $$S$$ is discrete, we can find a set $$T$$ of open intervals such that each point of $$S$$ is in exactly one of the intervals, and no two intervals intersect.
2. For each interval in $$T$$ we can pick one rational that is contained in it. Call this set of rationals $$R$$.
3. Because the elements of $$T$$ are pairwise disjoint, there is a one-to-one correspondence between $$T$$ and $$R$$, so $$|T|=|R|$$.
4. $$R$$ is a subset of the the rationals, so is at most countable; therefore $$T$$ is at most countable.
5. Elements of $$T$$ are in one-to-one correspondence with elements of $$S$$, so $$|S|=|T|$$.

The important thing to think about here is: what part of the proof fails when $$S$$ is not discrete? Say, when $$S = \{1, \frac12, \frac13, \frac14, \ldots, 0\}$$.

• what's the answer of your last question? I can find that in your set S, 0 is in fact a limit point of S. Hence, 0 is not isolated. – Mariana Jun 13 at 18:52
• As you say, $0$ is not isolated. Which of steps 1–5 fails? – MJD Jun 13 at 18:53
• I want to make a pedagogical point here. Answering a question like this for a student is not the same as preparing an answer to be handed in as homework. My answer skips over many details. Particularly in step 1: how do we know we can find the set $T$? How can we do this? This is really the crucial point. As a homework answer, my answer is inadequate. But my goal here is not to produce an acceptable homework answer. That is the student's job. The goal is to give the student a clear enough picture of the basic argument that they can make forward progress on filling in the details. – MJD Jun 13 at 18:55
• If S is not discrete, then we can not guarantee that "no two intervals intersect." – Mariana Jun 13 at 19:05
• @mariana Another thing you might ask yourself: the argument involves a set of intervals $T$.Could it be fixed up to work in a space like $\Bbb R^2$, where "intervals" does not make sense? And more generally: what are the necessary properties a space must have, if we want every discrete subset of the space to be countable? – MJD Jun 13 at 19:34

Intuitively, $$x$$ is an isolated point of $$S$$ if we can "draw some circle" around $$x$$ such that $$x$$ is the only point in that circle which is an element of $$S$$. This is because a point is isolated iff it is not a limit point, and $$x$$ is a limit point of $$S$$ iff we can find elements of $$S$$ (other than $$x$$ itself) arbitrarily close to $$x$$.

OK, so now suppose every element of $$S$$ is an isolated point of $$S$$. Imagine drawing a whole bunch of "isolating bubbles" around the elements of $$S$$. One thing we can try ot do is, for each $$x\in S$$, pick a rational $$q_x$$ in the "isolating bubble" around $$x$$. The idea then is that since no two elements of $$S$$ lie in the same "isolating bubble," the rationals we pick should be different.

However, this doesn't quite work. What if the "isolating bubbles" themselves overlap? E.g. consider $$S=\{0,1\}$$. Then the interval $$(-{2\over 3},{2\over 3})$$ is an "isolating bubble" around $$0$$, the interval $$({1\over 3}, {4\over 3})$$ is an "isolating bubble" around $$1$$, but the rational number $${1\over 2}$$ lies in both of those "isolating bubbles" so we might pick it for $$0$$ and $$1$$ simulatenusly.

What we need to do is pick really small "isolating bubbles" - so small that they're guaranteed to not overlap at all. In the case above, we could for example use $$(-{1\over 2}, {1\over 2})$$ as the "bubble" around $$0$$ and $$({1\over 2},{3\over 2})$$ as the "bubble" around $$1$$. One way to do this is to go back to the original "bubbles" we drew and shrink them a bit:

• Since $$S$$ is isolated, for each $$x\in S$$ we can find a $$\delta_x>0$$ such that $$(x-\delta_x,x+\delta_x)\cap S=\emptyset$$.

• Now let $$I_x$$ be the smaller interval $$(x-{\delta_x\over 2},x+{\delta_x\over 2})$$. Show that for $$x,y\in S$$ distinct we have $$I_x\cap I_y=\emptyset$$. So we've found isolating bubbles which don't overlap at all.

• Now apply the density of $$\mathbb{Q}$$: for each $$x\in S$$ pick some rational number $$q_x\in I_x$$. Show that the map $$S\rightarrow\mathbb{Q}: x\mapsto q_x$$ is injective and so $$S$$ is countable (this is where the non-overlapping-ness of the $$I$$s comes in).

The tag "real-analysis" is added so I presume this is a discrete subset of the reals.

Let $$x\in S$$ be arbitrary. Then, by the hypothesis, we can choose $$\delta(x)$$ such that $$S\cap(x-\delta(x),x+\delta(x))=\emptyset$$ (i.e. isolated). "The rationals are dense in the reals" means that in any open set of reals, we can find a rational in it. For example $$(\pi - \frac{1}{n} ,\pi+\frac{1}{n})$$ contains a rational for all naturals $$n$$. Now choose a rational $$r(x)=\frac{p}{q}$$ in $$(x-\delta(x),x+\delta(x))$$ such that $$p+q$$ is minimal(the smallest in this set). If there are multiple of these such that $$p+q$$ is minimal, choose the one with the smallest $$p$$. This awkward choice of minimal is to avoid using AoC, we could otherwise just say "choose a rational in the neighbourhood". Now define $$f$$ to be a function such that $$f:S\rightarrow \mathbb{Q}$$, $$f(x)=r(x)$$. By nature of isolated points, this function is injective, thus $$|S|\sim |\mathbb{Q}|\sim |\mathbb{N}|$$

• Why $S\cap(x-\delta(x),x+\delta(x))=\emptyset$ ? I refer to Rudin definition 2.18 (c), it said if $p \in E$ and p is not a limit point of E, then p is called an isolated point of E. – Mariana Jun 13 at 18:35
• Your answer seems too far beyond my limited knowledge. Can you give an easier version? My knowledge is only from Baby Rudin Analysis textbook. – Mariana Jun 13 at 18:37
• @Mariana Re: your first comment, think about the contrapositive. If there is no $\delta>0$ such that $S\cap (x-\delta,x+\delta)=\emptyset$, do you see why that means that $x$ is a limit point of $S$? – Noah Schweber Jun 13 at 18:39
• This isn't actually a correct answer: what if $(x-\delta(x),x+\delta(x))\cap (y-\delta(y),y+\delta(y))\not=\emptyset$ for $x,y\in S$? You need the additional property that the "isolating intervals" don't overlap. – Noah Schweber Jun 13 at 18:42
• @NoahSchweber The real numbers are a Hausdorff space, this is not a problem. EDIT actually just need to set the radius to delta/2 – aristotlefromgreece Jun 14 at 13:35