Suppose we have non-empty [itex]A_{1}[/itex] and non-empty [itex]A_{2}[/itex] which are both open. By "open" I mean all points of [itex]A_{1}[/itex] and [itex]A_{2}[/itex] are internal points. There is an argument -- which I have seen online and in textbooks -- that [itex]A_{1} \cap A_{2} = A[/itex] is open (assuming [itex]A[/itex] is non-empty) since:(adsbygoogle = window.adsbygoogle || []).push({});

1. For some [itex]x \in A, A_{1}, A_{2}[/itex] we have neighborhoods [itex]N_{1}[/itex] and [itex]N_{2}[/itex] around [itex]x[/itex] of radii [itex]r_{1}[/itex] and [itex]r_{2}[/itex] respectively, such that [itex]N_{1} \subset A_{1}[/itex] and [itex]N_{2} \subset A_{2}[/itex]. We know this since [itex]A_{1}[/itex] and [itex]A_{2}[/itex] are stipulated to be open.

2. If we take [itex]min(r_{1},r_{2}) = r[/itex] we can then construct a neighborhood [itex]N[/itex] of radius [itex]r[/itex] around [itex]x[/itex] and (this is the part that seems false to me) we can somehow KNOW that [itex]N \subset A[/itex] and therefore conclude that [itex]A[/itex] is open.

But why (2)? How do we KNOW that [itex]N \subset A[/itex]? Why can't [itex]N_{1} \subset A_{1}[/itex] and [itex]N_{2} \subset A_{2}[/itex] but nevertheless [itex]N \not \subset A[/itex]?

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# Finite Intersection of Open Sets Are Always Open?

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