Basic Analysis – Gently Done Topological Vector Spaces

88 Basic Analysis
We shall apply this result to quotient spaces, to which we now turn. Let X be
a vector space, and let M be a vector subspace of X. We define an equivalence
relation ∼ on X by x ∼ y if and only if x−y ∈ M. It is straightforward to check
that this really is an equivalence relation on X. For x ∈ X, let [x] denote the
equivalence class containing the element x. X/M denotes the set of equivalence
classes. The definitions [x]+[y] = [x+y] and t[x] = [tx], for t ∈ K and x,y ∈ X,
make X/M into a linear space. (These definitions are meaningful since M is a
linear subspace of X. For example, if x ∼ x ′ and y ∼ y ′ , then x+y ∼ x ′ +y ′ , so
that the definition is independent of the particular representatives taken from the
various equivalence classes.) We consider the possibility of defining a norm on the
quotient space X/M. Set
�[x]� = inf{�y� : y ∈ [x]}.
Note that if y ∈ [x], then y ∼ x so that y −x ∈ M; that is, y = x+m for some
m ∈ M. Hence
�[x]� = inf{�y� : y ∈ [x]} = inf{�x+m� : m ∈ M}
= inf{�x−m� : m ∈ M},
where the last equality follows because M is a subspace. Thus �[x]� is the distance
between x and M in the usual metric space sense. The zero element of X/M is
[0] = M, and so �[x]� is the distance between x and the zero in X/M. Now, in a
normed space it is certainly true that the norm of an element is just the distance
between itself and zero; �z� = �z −0�. This suggests that the definition of �[x]�
above is perhaps a reasonable one.
To see whether this does give a norm or not we consider the various requirements.
First, suppose that t ∈ K, t �= 0, and consider
�t[x]� = �[tx]�
= inf
m∈M �tx+m�
= inf �tx+tm�, since t �= 0,
m∈M
= |t| inf
m∈M �x+m�
= |t|�[x]�.
If t = 0, this equality remains valid because [0] = M and inf m∈M �m� = 0.
Department of Mathematics King’s College, London