Measure, Integration & Real Analysis, 2021a

152 Chapter 6 Banach Spaces
Entrance to the École Polytechnique (Paris), where Augustin-Louis Cauchy
(1789–1857) was a student and a faculty member. Cauchy wrote almost 800
mathematics papers and the highly influential textbook Cours d’Analyse (published
in 1821), which greatly influenced the development of analysis.
CC-BY-SA NonOmnisMoriar
Every nonempty subset of a metric space is a metric space. Specifically, suppose
(V, d) is a metric space and U is a nonempty subset of V. Then restricting d to
U × U gives a metric on U. Unless stated otherwise, you should assume that the
metric on a subset is this restricted metric that the subset inherits from the bigger set.
Combining the two bullet points in the result below shows that a subset of a
complete metric space is complete if and only if it is closed.
6.16 connection between complete and closed
(a) A complete subset of a metric space is closed.
(b) A closed subset of a complete metric space is complete.
Proof We begin with a proof of (a). Suppose U is a complete subset of a metric
space V. Suppose f 1 , f 2 ,... is a sequence in U that converges to some g ∈ V.
Then f 1 , f 2 ,...is a Cauchy sequence in U (by 6.13). Hence by the completeness
of U, the sequence f 1 , f 2 ,... converges to some element of U, which must be g
(see Exercise 7). Hence g ∈ U. Now6.9(e) implies that U is a closed subset of V,
completing the proof of (a).
To prove (b), suppose U is a closed subset of a complete metric space V. To show
that U is complete, suppose f 1 , f 2 ,...is a Cauchy sequence in U. Then f 1 , f 2 ,...is
also a Cauchy sequence in V. By the completeness of V, this sequence converges to
some f ∈ V. Because U is closed, this implies that f ∈ U (see 6.9). Thus the Cauchy
sequence f 1 , f 2 ,...converges to an element of U, showing that U is complete. Hence
(b) has been proved.
Measure, Integration & Real Analysis, by Sheldon Axler