Cantor set nowhere dense but still perfect?

Question

So I want to prove that the Cantor metric space $(C,|\cdot|)$ is perfect (that is, closed and has no isolated point). However, $C$ is known to be nowhere dense in $\mathbb{R}$, which means that if $x\in\mathbb{R}$ then $\exists \epsilon > 0$ such that $\forall y\in C$, $|x-y|>\epsilon$. However, this appears to imply that all points of $C$ are isolated. But then $C$ would not be perfect.

So how can it be that a set is nowhere dense, and yet has no isolated points?

Answer 1

That definition of a nowhere dense set isn't right. The name itself of that term might be misleading, actually, but I'll try to go through a systematic line of reasoning to show what you're actually supposed to negate.

I apologize if this is too pedantic.

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First, I'll define what dense means and what "not dense" means below.

A set $S$ is dense in $\mathbb{R}$ if for every $x\in\mathbb{R}$ and $\epsilon > 0$ there is some $y\in S$ such that $|x - y| < \epsilon$.

I find it easier to think about it in the language of intervals here:

A set $S$ is dense in $\mathbb{R}$ if for every $x\in\mathbb{R}$, every open interval of the form $J = (x - \epsilon, x + \epsilon)$ intersects $S$.

Okay, now the negation of this is:

A set $S$ is not dense if for some $x\in\mathbb{R}$, there is an open interval of the form $J = (x - \epsilon, x + \epsilon)$ that is devoid of $S$.

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Now I'll make precise what might be meant by "somewhere dense" (this is not a real term though):

A set $S$ is "somewhere dense" if somewhere, on some open interval $I$, $S$ is dense on that place.

More formally,

A set $S$ is "somewhere dense" if there exists some open interval $I$ such that $S\cap I$ is dense in $I$.

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This is what we'd want to negate, and from this we have the following definition:

A set $S$ is nowhere dense if for all open intervals $I$, $S\cap I$ is not dense in $I$.

Now at this point, we expand upon what is meant by "not dense":

A set $S$ is nowhere dense if for all open intervals $I$, there exists some $x\in I$ such that there is an interval of the form $J = (x - \epsilon, x + \epsilon)$ that is devoid of $S$.

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What this means is that no matter where you look -- no matter what open interval $I$ you look at -- you can find an interval $J\subseteq I$ somewhere there that doesn't intersect $S$.

If I pick a point $z\in\mathbb{R}$, then I can look around that point by examining intervals of the form $I = (z - \epsilon, z + \epsilon)$. For every such interval, I can find a $J\subseteq I$ that does not intersect $S$. I can't be sure where $J$ will be ahead of time, but it will be there somewhere. Note that $J$ doesn't even have to contain $z$; it just needs to be somewhere in $I$. Now If I keep decreasing $\epsilon > 0$, then that means I will keep finding $J\subseteq \mathbb{R} - S$ that seem to get "closer and closer" to $z$ (by being bound by $I$).

Now how does this apply to the Cantor set $C$? The set is closed and it has no isolated point because every $z\in C$ has a sequence of points $z_{1}, z_{2},\ldots$ from $C - \{z\}$ that approach $z$. But at the same time, you can find a sequence of intervals $J_{1}, J_{2}, \ldots$ that don't intersect $C$ but they also keep getting "closer and closer" to $z$.

When we view it from this perspective, it manages to maintain both properties without any contradiction.