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.
Best Answer
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 $(x-\delta_x,x+\delta_x)\cap S=\{x\}$.
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).