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I'm a beginner in Topology. Today, this came up my mind:

(1) For a set $X$, choose a subset $A\subseteq X$. Let $S\subseteq X$ be a closed set if and only if $(A\subseteq S)\vee (S=\emptyset)$. This is a topology on $X$.

(2) For a set $X$, let $S\subseteq X$ be a closed set if and only if $(S $ is finite$)\vee (S = X)$. This is another topology on $X$.

The questions are:

[1] What are these two topologies called?

[2] Do they have significance? Are they used somewhere?

First question on stackexchange. Thanks!

twelvefaces
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    I haven't heard about the first one, the second one is the cofinite topology which is in particular useful for constructing counterexamples – SBF Jan 20 '13 at 12:02
  • "Cofinite topology occurs in the context of the Zariski topology." (http://en.wikipedia.org/wiki/Cofiniteness#Cofinite_topology) It seems useful in Algebraic Geometry. – Karatuğ Ozan Bircan Jan 20 '13 at 12:10
  • For topology (1), the connected sets are precisely the empty set, singletons and every set that intersects $A$. So, could we see $A$ "complete subgraph", where every pair of points in $A$ is connected, and $A$ is the center of a "star graph", where every point in $A$ is connected to any point of $X$? – Herng Yi Jan 20 '13 at 12:17

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Your second topology is known as the cofinite topology on $X$; it’s important because a topology $\tau$ on $X$ is $T_1$ if and only if $\tau$ contains the cofinite topology. As Ilya notes in the comments, it is also useful in the construction of counterexamples, at least when the examples are not required to be Hausdorff.

I don’t know of any name for your first topology. It’s the discrete topology on $X\setminus A$ together with the indigestible lump $A$, a set that behaves in most respects like a single point. The quotient space $X/A$ is simply a discrete space with one point corresponding to $A$ and one point for each point of $X\setminus A$.

Brian M. Scott
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    Would it be correct to say that the first topology is the direct sum of the discrete topology on $X \setminus A$ and the indiscrete topology on $A$? – Herng Yi Jan 20 '13 at 12:29
  • @HerngYi: Yes, it would (though I’m more accustomed to calling it the disjoint union). – Brian M. Scott Jan 20 '13 at 12:30
  • @Brian: Isn't the the disjoint union $A\bigsqcup X\backslash A$ finer than the first topology? Among the closed sets in the union there are all subsets of $X\backslash A$, disjoint from $A$. But those sets are not closed in the first topology. – Stefan Hamcke Jan 20 '13 at 14:04
  • @Stefan: You’re right; that was the last thing I did before going to bed, and it shows. Thanks for catching it. – Brian M. Scott Jan 20 '13 at 19:50
  • @HerngYi: Please ignore my answer to your question: it’s wrong. Stefan is correct: the direct sum of $A$ (indiscrete) and $X\setminus A$ (discrete) would have $X\setminus A$ as an open set, and your topology doesn’t. – Brian M. Scott Jan 20 '13 at 19:52