# Is it always true that the complement of a closed set is open?

In a metric space $M$, by definition, a subset $F$ is closed if $M\setminus F$ is open.

However in a general topological space $T$, say with topology $\mathcal{T}$ is this always true? For example with the Zariski topology $\mathcal{T}_z$ defined: Closed sets consist of finite subsets of $T$, also $T$ is closed. Does it still hold that for some closed set $F \subset T$ that $T\setminus F$ is open?

I also ask the equivalent statement for open sets, namely if $U$ is open is $T\setminus U$ necessarily closed?

• Yes. This is how closed sets are defined. – David Mitra May 14 '15 at 9:53
• One thing to always remember is that just because a set is closed doesn't mean it's not open. In $\mathbb{R}$ clopen sets are quite rare but in general topological spaces this doesn't need to be the case. – DRF May 14 '15 at 9:58
• This reminds me of my lecturer in General Topology. He was quite an ambiguous fellow, and was rarely clear or precise about what he said. When introducing the concept of an open/closed set he told us "The door parable", which was: "A set is like a door. It can be either open, or closed. Except that a set can also be neither. Oh, and also both. You know what, forget about it..." – yohBS May 14 '15 at 10:31

Yes, that's always true. In general topology, you define the closed sets to be the complements of open sets. A topology $\tau$ on a set $X$ is just a subset of the power set of $X$ with certain properties, whose elements are called "open sets" and whose complements are called "closed sets".
You can define closed sets as sets that contain all of their limit points and also get the theorem "$A$ is open $\iff$ $A^{c}$ is closed". Which I think is a more interesting way to present the material than just defining closed as complement of open.