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This is from Rudin's Principles of Mathematical Analysis, Example 2.21.

Let us consider the following subset of $R^2$:

(c) A nonempty finite set.

The book says it is closed and bounded but not open nor perfect.

My problem is with it being closed, I don't understand why. I know there is already a similar post for this here, but I don't quite get it.

I know that a closed set contains all its limit points. I imagined a set $S$ of discrete points on the plane. For each point, if we construct a radius, we find that the $N_r(p)$ for each p in the set will not contain any points of S. So I don't think any of them are limit points.

But then if I take a look at its complement, it is clear that the complement $S^c$ is open which means that the set is closed. So I am confused. Also, I think I remember something about the definition of a closed set being vacuously true but I'm not sure how to relate it here (I lost my notes for this).

So my questions would be:

1) Why is the set closed (without using the complement of $S$, possibly using logic to think it through)?

2) Is my reasoning using the complement of $S$ correct?

Thank you.

August
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    If a set has no limit points at all, it vacuously contains all of its limit points. – Daniel Fischer Oct 27 '14 at 00:30
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    Both of your points are correct ways of proving that finite sets are closed. If you wonder why it's closed in the sense of satisfying the definition, then you've already provided the answer. If you want the intuitive reason they are closed, that's because you cannot take the limit of a sequence in a finite set and end up outside that set. Intuitively, (good falsehood) open sets are like open doors - if you run fast enough you can get out the door. Closed sets are like closed doors, you cannot leave. – J. David Taylor Oct 27 '14 at 00:45

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