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.

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    $\begingroup$ What part don't you understand? $\endgroup$ – MJD Jun 13 at 18:26
  • $\begingroup$ See for example math.stackexchange.com/q/961689/42969 $\endgroup$ – Martin R Jun 13 at 18:27
  • $\begingroup$ I don't understand the isolation of each point + rationals are dense part. $\endgroup$ – Mariana Jun 13 at 18:28
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    $\begingroup$ 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: $d(x,y)=\mathbb{1}(x\neq y)$). $\endgroup$ – Oliver Diaz Jun 13 at 18:38
  • $\begingroup$ If the dimension is infinite, for example the set of real sequences, it is not true. Take the subset of integer sequences. $\endgroup$ – 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 $S = \{1, \frac12, \frac13, \frac14, \ldots, 0\}$.

  • $\begingroup$ 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. $\endgroup$ – Mariana Jun 13 at 18:52
  • $\begingroup$ As you say, $0$ is not isolated. Which of steps 1–5 fails? $\endgroup$ – MJD Jun 13 at 18:53
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    $\begingroup$ 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. $\endgroup$ – MJD Jun 13 at 18:55
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    $\begingroup$ If S is not discrete, then we can not guarantee that "no two intervals intersect." $\endgroup$ – Mariana Jun 13 at 19:05
  • $\begingroup$ @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? $\endgroup$ – 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 $|S|\sim |\mathbb{Q}|\sim |\mathbb{N}|$

  • $\begingroup$ 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. $\endgroup$ – Mariana Jun 13 at 18:35
  • $\begingroup$ Your answer seems too far beyond my limited knowledge. Can you give an easier version? My knowledge is only from Baby Rudin Analysis textbook. $\endgroup$ – Mariana Jun 13 at 18:37
  • $\begingroup$ @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 $S$? $\endgroup$ – Noah Schweber Jun 13 at 18:39
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    $\begingroup$ 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. $\endgroup$ – Noah Schweber Jun 13 at 18:42
  • $\begingroup$ @NoahSchweber The real numbers are a Hausdorff space, this is not a problem. EDIT actually just need to set the radius to delta/2 $\endgroup$ – aristotlefromgreece Jun 14 at 13:35

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