# Thread: Clarify def of closed set: implication?

1. ## Clarify def of closed set: implication?

Is the definition of a closed set an implication?

My understanding of the definition is this:

OPEN: For a set U in R. U is open if for all x in U there exists an m>0 s.t. (x-m, x+m) is also in U.

CLOSED: For a set F=R\U. If F is open then U is closed.

So if I find that F is not open, does that mean that U is not closed?

Thanks a bunch!

2. Hello yvonnehr!

Originally Posted by yvonnehr
Is the definition of a closed set an implication?

My understanding of the definition is this:

OPEN: For a set U in R. U is open if for all x in U there exists an m>0 s.t. (x-m, x+m) is also in U.
Yes, that's right

Originally Posted by yvonnehr
CLOSED: For a set F=R\U. If F is open then U is closed.

So if I find that F is not open, does that mean that U is not closed?
If F is open, then F^c is closed (^c is supposed to mean 'complement of')

So if F is not open, I assume F is closed. What is F^c? It is U. So U is open.

But: R and $\displaystyle \emptyset$ is both open and closed.

Kind regards
Rapha

3. ## Alternative definition

You might also find it helpful to know that you can characterize closed a set by knowledge of the limits of its sequences. Let $\displaystyle (X,d)$ be a metric space, then $\displaystyle F\subset X$ is closed if and only if, whenever the sequence $\displaystyle (x_{n})_{n\in\mathbb{N}} \in X$ converges - to $\displaystyle x,$ say, - then it follows that $\displaystyle x\in F$. I.e.: $\displaystyle F$ is closed if and only if it contains all of its limits. You can prove this using your definition.

4. Originally Posted by Rapha
So if F is not open, I assume F is closed. What is F^c? It is U. So U is open.
Be very careful with statements such as that.
A door is either open or closed. That ain't true for sets.
The set $\displaystyle [0,1)$ is neither open nor closed.
Its complement $\displaystyle (-\infty,0)\cup [1,\infty)$ is also neither open nor closed

If is true that a set is closed in and only if its complement is open.

As has already been pointed out, the standard metric space definition of closed sets is:
A set is closed if and only if it contains all of its limit points.

5. ## closed set-biconditional statement

Originally Posted by Plato
Be very careful with statements such as that.
A door is either open or closed. That ain't true for sets.
The set $\displaystyle [0,1)$ is neither open nor closed.
Its complement $\displaystyle (-\infty,0)\cup [1,\infty)$ is also neither open nor closed

If is true that a set is closed in and only if its complement is open.

As has already been pointed out, the standard metric space definition of closed sets is:
A set is closed if and only if it contains all of its limit points.
This makes sense to me. So, the definition is a biconditional statement rather than an implication. I just wanted to make sure so that I can use is properly in a proof. Thank you so much, Plato.