Topology Defined by Closed Sets

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Theorem

Let $X$ be any set and let $\tau$ be a collection of subsets of $X$.

Then $\tau$ is a topology on $X$ if and only if:

$(1): \quad$ Any intersection of arbitrarily many closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$
$(2): \quad$ The union of any finite number of closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$
$(3): \quad X$ and $\varnothing$ are both closed sets of $X$ under $\tau$.


Proof

From the definition, if $V$ is a closed set of $X$, then $X \setminus V$ is an open set of $X$.

Let $\mathbb V$ be any arbitrary set of closed sets of $X$.

Then by De Morgan's Laws: Difference with Intersection, we have:

$\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \left({X \setminus V}\right)$


First, let $\tau$ be a topology on $X$.

We have that:

Intersection of Closed Sets is Closed
Finite Union of Closed Sets is Closed
By Open and Closed Sets in Topological Space, $\varnothing$ and $X$ are both closed in $X$.

Thus the properties as listed above hold.

$\Box$


Now, suppose the properties:

$(1): \quad$ Any intersection of arbitrarily many closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$
$(2): \quad$ The union of any finite number of closed sets of $X$ under $\tau$ is a closed set of $X$ under $\tau$
$(3): \quad X$ and $\varnothing$ are both closed sets of $X$ under $\tau$.

all hold.

That means $\displaystyle \bigcap \mathbb V$ is closed.

So $\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \left({X \setminus V}\right)$ is open.

Thus we have that the union of arbitrarily many open sets of $X$ under $\tau$ is an open set of $X$ under $\tau$.

Similarly we deduce that the intersection of any finite number of open sets of $X$ under $\tau$ is an open set of $X$ under $\tau$.

And of course by Open and Closed Sets in Topological Space, $\varnothing$ and $X$ are both open in $X$.

So $\tau$ is a topology on $X$.

$\blacksquare$


Sources