Why the definition of a topological space defined under finite intersection and arbitrary union What if we change the conditions by arbitrary intersection and finite union?
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General topology arose in large part as a generalization of real analysis. On the real line with the usual topology, an arbitrary union or finite intersection of open intervals is open.
We cannot push to even countably infinite intersections, however, since $$ \bigcap_{n=1}^\infty \left(-\frac{1}{n}, \frac{1}{n} \right) = \{0\} $$ gives us an example of a countably infinite collection of open intervals whose intersection is not open.
Austin Mohr
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While defining topology in general, mathematicians wanted to generalize the understanding of "openess" from metric spaces such as $\mathbb{R}^N$, defined by the means of open balls.
Thus, for example, the intersection of sets of the form $(-\frac{1}{n},\frac{1}{n})$ (where $n$ is any integer) should not be open, since it equals a singleton point $0,$ which is not open in the metric topology.
Similarly, the union of sets of the form $(-r,r)$ (where $r$ is any real number) should be open, since the union equals $\mathbb{R}$, which is open in metric topology.
Pawel
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