# Open, closed set in metric space [closed]

I started to study about Metric space at uni and am confused with the definitions open and closed set. It seems to me that being an open set always satisfies the definition of a closed set.

## closed as unclear what you're asking by Parcly Taxel, Namaste, Jack, TheSimpliFire, SaadFeb 25 '18 at 9:37

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• Could you state those definitions? – MPW Feb 24 '18 at 2:46
• In my note, the definition of a closed set says, U is closed precisely when Bε(x) ∪ U non-empty for every ε>0 – Alexander Denis Feb 24 '18 at 2:51
• What you have written down is not the definition of a closed set. It looks like a garbled version of the definition of a point $x$ being in the closure of a set $U$. You need to start by looking up the correct definitions. – fredgoodman Feb 24 '18 at 3:02
• What is your question? – Jack Feb 25 '18 at 2:55

A topology $\tau$ on a space $X$ is just a collection of sets which are specified as being open (by definition)...

When $X$ is a metric space, a set $U$ being open is equivalent to the existence of an epsilon ball, $B(x,\epsilon)\subset U$ for any $x\in U$, centered at $x$ and contained in $U$...

Always, for any topological space (whether metric or not), the complement of an open set is defined to be closed ... It follows immediately that the complement of a closed set is open...

In metric topology every point in an open set is the center of an open ball which is contained in the open set.

That is points come with an open ball around them.

On the other hand a set is closed if its complement is open.

Thus the complement of open sets are closed sets and the complement of closed sets are open sets.

For example the open interval $(-1,1)$ is open and its complement $$(-\infty , -1] \cup [1, \infty )$$ is closed.

The closed interval $[-1,1]$ is closed and its complement $$(-\infty , -1) \cup (1, \infty )$$ is open.