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I simply cannot wrap my head around the fact that the Cantor set has neither interior points nor isolated points.

As I understand it, an interior point $a\in A$ is one such that there is at least one neighborhood of $a$ that is a subset of $A$. Moreover, an isolated point $b\in B$ is one such that there is at least one neighborhood of $b$ that is disjoint of $B$.

How can the Cantor set fit both criterion?

If the Cantor set $C$ has no isolated points, then a point $c\in C$ must be a limit point. That is, every $\varepsilon$-neighborhood of $c$ intersects $C$ in some point different from $c$. Furthermore, if the Cantor set has no interior points, then every $\varepsilon$-neighborhood of $c\in C$ must not be a subset of $C$. Somehow, putting these definitions together does not seem to help me "see" this in my mind since it is so counter-intuitive.

Does anyone know of a more "natural" way to view this? Also, would $\mathbb{Q}\subset\mathbb{R}$ fit this criteria as well?

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Yes: $\Bbb Q$ as a subset of $\Bbb R$ has empty interior and no isolated points. The same is true of $\Bbb R\setminus \Bbb Q$, the set of irrationals. –  Brian M. Scott May 8 '12 at 3:29

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Let $A$ be a set. Interior points of $A$ are points $a\in A$ for which there exists some neighborhood $U$ of $a$ such that for all $x\in U$, $x\in A$. On the other hand, isolated points of $A$ are points $a\in A$ for which there exists some neighborhood $U$ of $a$ such that for all $x\in U$ other than $a$, $x\notin A$. In a sense, they are two opposite extremes. A point can easily be neither: we need only that for all neighborhoods $U$ of $a$, there is some point in $U$ other than $a$ which is in $A$, yet not all points in $U$ are in $A$.

A simple example is the point $0$ in the set $[0,1]$. This is not an interior point of $A$, as any neighborhood of $0$ contains negative numbers, but is not isolated either, as any neighborhood of $0$ contains numbers in $[0,1]$. A similar but slightly more interesting example is the set $\{0\}\cup \{1,1/2,1/3,\ldots\}$, in which $0$ is also neither an interior nor an isolated point.

As you suspected, all points in $\mathbb Q$ are neither interior nor isolated, as for any $a\in \mathbb Q$ we have rational and irrational points distinct from but arbitrarily close to $a$ thus any neighborhood of $a$ contains distinct points in $\mathbb Q$ and not in $\mathbb Q$.

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The last sentence in your first paragraph did it for me! I believe I understand it now, although your example with $V=\{0\}\cup\{1,1/2,1/3,\dots\}$ left me completely perplexed! How can $0$ be an isolated point in this case? I cannot seem to find a neighborhood $U$ of $0$ (which excludes $0$) such that $U\cap V=\varnothing$. –  Josué Molina May 8 '12 at 3:56
    
@JosuéMolina That was a mistake, I meant to write "also neither an interior nor an isolated point". –  Alex Becker May 8 '12 at 4:26

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