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Theorem A set E is open if and only if the complement $E^{c}$ is closed

Proof

$E$ open $\iff$ any point $x \in E$ is an interior point

$\iff \forall x \in E, \exists$ a neighborhood $N$ of x s.t. $N$ is disjoint from $E^{c}$

$\iff \forall x\in E$, $x$ is not a limit point of $E^{c}$

$\iff E^{c}$ contains all its limit point

Here is my question for the last justification. How does showing that $x$ is not a limit point of $E^{c}$ imply that the $E^c$ contains all the limit points?

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4 Answers 4

up vote 5 down vote accepted

Suppose that $E^c$ had a limit point $x$ that was not in $E^c$; then $x$ would belong to $E$ and therefore would not in fact be a limit point of $E^c$.

In other words, if nothing in $E$ is a limit point of $E^c$, then any limit points that $E^c$ might have can’t be in $E$ and must therefore be in $E^c$. (Remember, anything that’s not in $E$ is in $E^c$ and vice versa.)

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If every $x\in E$ is not a limit point of $E^{c}$, then no limit point of $E^{c}$ can be found in $E$; otherwise, there would be some $x\in E$ which is a limit point of $E^{c}$, but we have already shown this to be false. So, since no limit point of $E^{c}$ can be found in $E$, it follows that every limit point of $E^{c}$ is in $E^{c}$. So, $E^{c}$ contains all of its limit points as desired and $E^{c}$ is closed.

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All the limit points of $E^\rm{c}$ are either in $E^\rm{c}$ or in $E$. You've shown that points in $E$ cannot be limit points so that can only mean that $E^\rm{c}$ contains all it's limit points.

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All the limit points of $E^c$ are either in $E^c$ or in E. Huh is this really true? How could a limit point of a set be in its complement? I am talking about the complement of $E^c$, which is $(E^c)^c = E$ –  sidht Nov 4 '12 at 1:30
    
Consider a sequence approaching the boundary of the open unit disk. The limit point itself is contained in the unit circle which is in the complement. Another way to see this is to simply note that $E^\rm{c}$ and $E$ together contains every point (by definition of complement). –  EuYu Nov 4 '12 at 1:32

The following two statements are equivalent (they are each other's contrapositive):

If $x\in E$, then $x$ is not a limit point of $E^c$. (A rephrasing of $\forall x\in E$, $x$ is not a limit point of $E^c$.)

If $x$ is a limit point of $E^c$, then $x\notin E$ (that is $x\in E^c$).

That's how you make that last leap.

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