Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 11: Spaces of functions, uniform convergence

Convergence and boundedness

Let $(X,d)$ be a metric space.

Let $A\subseteq X\text{.}$

The set $A$ is bounded if $A$ satisfies: there exists $M\in {ℝ}_{\ge 0}$ such that
if $a_1,a_2\in A$ then $d(a_1,a_2)\le M\text{.}$

HW: Define the diameter of $A,$ the distance between $A$ and $B,$ and the distance between $x$ and $A\text{.}$

Convergence and boundedness

HW: Let $(X,d)$ be a metric space and let $$\begin{array}{cccc}\stackrel{\rightharpoonup }{x}:& {\mathbb{Z}}_{>0}& ⟶& X\\ & n& ⟼& x_n\end{array}$$ be a sequence in $X\text{.}$ Show that if $\stackrel{\rightharpoonup }{x}$ converges then $\{x_1,x_2,x_3,\dots \}$ is bounded.

		Proof.
	Assume $\stackrel{\rightharpoonup }{x}$ converges. To show $\{x_1,x_2,\dots \}$ is bounded. We know: There exists $x\in X$ such that $\underset{n\to \infty }{\text{lim}}{x}_n=x\text{.}$ To show: There exists $M\in {ℝ}_{\ge 0}$ such that if $i,j\in {\mathbb{Z}}_{>0}$ then $d(x_i,x_j)\le M\text{.}$ Let $N\in {\mathbb{Z}}_{>0}$ be such that if $n\in {\mathbb{Z}}_{>0}$ and $n\ge N$ then $d(x_n,x)<1\text{.}$ Let $M=2\text{max}\{1,d(x_1,x),\dots ,d(x_N,x)\}\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 11 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 14, 2014.

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