Let S be a normed vector space and let U be a linear subspace. Build a pseudo norm on S as the distance from x to U. Technically, this is the greatest lower bound of the distances from x to all the points of U. We need to show this is a pseudo norm.
If q is the point in U with minimum distance to p, scale p and q by c and the distance is multiplied by c. Yet the distance to every other point in U is also multiplied by c. (Everything in U is c times something else in U.) Thus q remains the closest point, and distance is scaled by c. If there is no minimum q, let q be a sequence of points whose distance from p approaches the lower bound. All distances are multiplied by c, and the lower bound is multiplied by c, as illustrated by c*q.
The triangular inequality is inherited from S. Let x and y be any two points in S, and let p and q be points in U that hold the distances |x,p| and |y,q| to within ½ε of their true distances from U. The distance from p+q to x+y is now bounded by the sum of the true distances to U, + ε. Let ε approach 0, and p+q proves the norm of x+y is no larger than the norm of x plus the norm of y. We have satisfied the properties of a pseudo norm.
If p is in the closure of U then a sequence q approaches p, and p is 0 distance from U. Extend U to the closure of U; hence U is a closed set in S. Verify that this does not change the distance from x to U.
With U closed, a point not in U is a positive distance from U. Otherwise a sequence of points in U approaches p, and p would be included in U. Thus U, and only U, has norm 0.
If y is in U, add y to x. This does not change the set of distances from x to the points of U. The distance to p has become the distance to p+y, and so on. Thus the shifted subspace x+U is a fixed distance from U, and has a well defined norm.
Collapse the points that have the same norm, giving a quotient space S/U. This is a topological quotient space, as described in the previous section, but it is also a linear quotient, where U is the kernle, and cosets of U build the image space. No ambiguity here; S/U is a quotient space.
If S is complete, is S/U complete?
For starters, U is closed by assumption, and a closed subspace of a complete metric space remains complete. If a cauchy sequence in U converges to p outside of U, then p is a positive distance from U, a positive distance from the sequence, and it cannot be the limit point of the sequence - so U is complete.
Use this fact to build a function f(x) that projects x onto the point in U that is closest to x.
Note that x-f(x) is actually the same point in S/U.
Let q be a sequence in S that becomes cauchy in S/U. Replace qi with qi-f(qi), which does not change its image in S/U. Now each point in the sequence q is closest to 0 in U, and the distance to U is simply qi. In other words, qi maps to qi in S/U. Let p be the limit of q in S. Suppose p is closest to z in U, where z is not 0. Far down the sequence, qi and p are less than ½|z| apart. Now p is closer to 0 than z. This is a contradiction, hence p is closest to 0 in U, and p maps to p in S/U. The limit exists, namely p, and the quotient space is complete.
Given a banach space, mod out by a closed linear subspace and find another banach space.