# Proving a set is closed

What is need to be proven, in a proof by contradiction that a set is closed?

If we have to show that $K$ is closed that mean that we need to show that $K^{C}$ is open. Let there be $x\in K^{C}$ we need to show that $B(x,\delta)\subset K^{C}$

What is the contrary assumption?

For all $\delta>0; B(x,\delta)\subset K$? or is it For all $\delta>0; B(x,\delta)\cap K\neq \emptyset$?

• Yes, the later. The statement K^c is open is "there exists a delta so that B is a subset of K^c". So the negation is "for no delta is B a subset of K^c" which is to say "for every delta B contains points not in K^c" which is the same as "for every delta B contains points in K". The former says for all delta B is entirely in K which if you draw pictures you can convince yourself isn't nesc. true. (It's a statement that is too strong. So strong it is probably not true.) – fleablood Nov 22 '16 at 19:09
• Notice "for all $\delta > 0 B(x,\delta) \subset K$ implies that for all $y \in X$, the entire metric space, if we let $\delta > d(x,y)$ then $y \in B(x, \delta) \subset K$. So $X \subset K$. So $K$ is the entire metric space!!!!! So... that is not the correct negation. – fleablood Nov 22 '16 at 19:57

$\exists x \notin K, \forall \delta>0, B(x, \delta) \cap K \neq \emptyset$

In any topological space $X$, the set $K \subset X$ is not closed if there exists a limit point of $K$ that isn't in $K$. (This is the negation of "$K$ is closed if it contains all its limit points.") For a metric space (in terms of balls), this means that there exists $x \in X\setminus K$ such that every open ball around $x$ intersects $K$: $$\exists x \in X \setminus K : \forall \delta > 0 : B(x,\delta) \cap K \neq \emptyset.$$

• What does the intersection means in the term of balls and limit points? – gbox Nov 22 '16 at 19:03
• This is true but I think the OP's definition of closed is "the complement is open". By really understanding definitions it's possible (and very important that every student do so at some point) to see that in metric spaces "closed means complement is open" and "closed means contains all limit points" are one and the same. But as the OP is working with the "compliment is open" def, this may come as a bit of an abstraction shift. I dunno. I hate the "compliment is open" definition personally. – fleablood Nov 22 '16 at 19:18
• "What does the intersection means in the term of balls and limit points?" If p is a limit point, it mean for all $\delta$, $B(p,\delta)$ contains a point of K that is not p. So $B(x,\delta)\cap K \ne \emptyset$ and $B(x,\delta)\cap K \ne =\{p\}$. Note $p$ may or may not be in K itself. But if $p$ isn't in K then K is not closed as for closed sets all limit points are in the set. And if all limit points are in the set, the set is closed. – fleablood Nov 22 '16 at 19:25

"What is need to be proven, in a proof by contradiction that a set is closed?"

There are two possible approaches depending on your definition of "closed":

1) $K$ is closed if $K^c$ is open.

So a proof by contradiction is to assume $K^c$ is not open.

The definition of "open" is: For all $x \in K^c$ there is some $\delta > 0$ so that $B(x,\delta) \subset K^c$.

The negation of that is: There is some $x \in K^c$ where no $\delta$ exists so that $B(x,\delta) \subset K^c$ which put another way:

There is some $x \in K^c$ where for all $\delta$ it is that $B(x, \delta) \not \subset K^c$ or in other words:

There is some $x \in K^c$ there for all $\delta$, $B(x, \delta)$ contains point that are not in $K^c$. i.e $B(x, \delta)$ contains points that are in $K$.

Or in other words:

There is some $x \not \in K$ so that for all $\delta$, $B(x, \delta) \cap K \ne \emptyset$.

Notice that "for all $\delta$, $B(x, \delta) \cap K \ne \emptyset$" and "for all $\delta$, $B(x, \delta) \cap K \ne \{x\}$" is the definition of "$x$ is a limit point of $K$".

2) $K$ is closed if all the limit points of $K$ are in $K$.

So a proof by contradiction is to assume there exists a limit point $x$ of $K$ so that $x \not \in K$.

The definition of limit point is: $x$ is a limit point if, for all $\delta > 0$ then $B(x,\delta)$ contains a point of $K$ that is not equal to $x$.

So $x \not \in K$ and for all $\delta$, $B(x, \delta)\cap K \ne \emptyset$ and $B(x, \delta)\cap K \ne \{x\}$.

Since $x \not \in K$ that second half of that is obvious ($x \not \in K$ so $x \not \in B(x, \delta) \cap K$ so $B(x, \delta) \cap K \ne \{x\}$) and we only need to assume

There is some $x \not \in K$ so that for all $\delta$, $B(x, \delta) \cap K \ne \emptyset$.

Notice, this is the exact same assumption for either definition. This is because the two definitions are equivalent.

Once you get down to the definitions of "open" and "limit point" the theorem: $K^c$ is open if and only if every limit point of $K$ is a point of $K$, is simply a matter of definition. So we may define "closed" as either "its complement is open" or as "contains all its limit points".