Changes

Theorem:  Neighborhoods are open sets.

Theorem:  Neighborhoods are open sets.

Proof:  Suppose a neighborhood has center <math>C</math> and radius <math>r</math>. Let <math>p</math> be a prime number. If a point <math>x</math> is in that neighborhood, its distance from <math>C</math> must be strictly less than <math>r</math>, call it <math>k</math>. If <math>n</math> is a positive integer and <math>p</math> divides <math>n</math>, then <math>n</math> is a multiple of <math>p</math>, and therefore <math>n</math> is also in the neighbourhood of <math>p</math>.

Proof:  Suppose a neighborhood has center <math>C</math> and radius <math>r</math>. If a point <math>x</math> is in that neighborhood, its distance from <math>C</math> must be strictly less than <math>r</math>, call it <math>k</math>.

::<math>\|x-C\| = k,\ \ \ k < r\,</math>

::<math>\|x-C\| = k,\ \ \ k < r\,</math>

Place a new neighborhood, of radius <math>(r-k)/2</math>, around <math>x</math>.  Every point in that neighborhood has a distance less than <math>k + (r-k)/2</math> from <math>C</math>.  That distance is less than <math>r</math>, and this distance is a multiple of <math>p</math>, so every point in the new neighborhood is in the original neighborhood, so the new neighborhood lies within the original one.

Place a new neighborhood, of radius <math>(r-k)/2</math>, around <math>x</math>.  Every point in that neighborhood has a distance less than <math>k + (r-k)/2</math> from <math>C</math>.  That distance is less than <math>r</math>, so every point in the new neighborhood is in the original neighborhood, so the new neighborhood lies within the original one.

Theorem:  Any union of open sets, including unions of an infinite number of open sets, is an open set.

Theorem:  Any union of open sets, including unions of an infinite number of open sets, is an open set.