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I'm learning real analysis.

I found that there are two classification of points: interior/exterior/boundary point and limit point.

What's the relationship between interior/exterior/boundary point and limit point ?

Jill Clover
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2 Answers2

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As an exercise (which should simultaneously answer your questions), prove the following statements:

An interior point cannot be an exterior point.

An exterior point cannot be an interior point.

A boundary point is neither an interior point nor an exterior point.

An exterior point is not a limit point.

An interior point can be a limit point.

Let $S$ be a set. Every boundary point of $S$ is a limit point of $S$ and its complement. (This statement is false if you define a limit point of $S$ to be a point $p$ so that every neighborhood of $p$ contains some $x\in S$, $x\neq p$. But if you allow $x = p$ in the definition then the statement is true.)

These are all trivial, some may be very trivial depending on what the definitions of these terms are for you.

Gyu Eun Lee
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    The last statement "Every boundary point of S is a limit point of S" may be incorrect. – FreshAir Jul 30 '15 at 07:10
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    @FreshAir That depends on your definition of limit point - the statement is correct if the definition is so that every point of $S$ is a limit point of $S$. But I think that's actually the less standard definition, so I should probably mention that. Thanks for the catch. – Gyu Eun Lee Aug 01 '15 at 02:39
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From a topological perspective you can say the following: (Assuming $S$ is the subset of a topological space $X$ relative to which you want to define interior/exterior/limit and $S^c$ denotes $S$'s complement)

  • To every interior point $i$ there is an open subset of $S$ containing $i$.
  • To every exterior point $e$ there is an open subset of $S^c$ containing $e$.
  • If an open subset of $S$ contains a boundary point $b$ then that subset contains both points of $S$ and of $S^c$. (Boundary points can be in either $S$ or $S^c$)
  • Neighborhoods of limit points $l$ always contain at least one point that is in $S$ and is not $l$ itself.

The interior, the boundary and the exterior are a (disjunct) partition of $X$.

Note the difference between the boundary and the limit points: Isolated points (points in $S$ for which holds: there exists an open subset of $X$ that contain the point as sole member of $S$) are in $S$, in the boundary of $S$ but are not limit points of $S$. The Venn-Diagram here shows this nicely.

Stefan Hansen
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user1965813
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