Complete metric space

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Given a metric space [ilmath](X,d)[/ilmath], if every Cauchy sequence converges to a limit (sequence) within [ilmath]X[/ilmath] then [ilmath]X[/ilmath] is a complete metric space[1][2]. That is to say:

  • Given a sequence [ilmath](x_n)_{n=1}^\infty[/ilmath], it converging to a limit [ilmath]x\in X[/ilmath] or being a Cauchy sequence are equivalent. Or in symbols:
  • [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(x_n,x)]\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n,m\in\mathbb{N}[n\ge m>N\implies d(x_n,x_m)<\epsilon][/ilmath]


Complete space

  • Obviously the [ilmath]\mathbb{R} [/ilmath] (reals) are complete, considered with the usual topology induced by the Absolute value metric

Incomplete space

  • A good example is the space of fractions, [ilmath]\mathbb{Q} [/ilmath] considered with the Absolute value metric again, there are rational sequences which converge to say, [ilmath]\sqrt{2} [/ilmath], and [ilmath]\sqrt{2}\notin\mathbb{Q} [/ilmath]
  • A better example is the space of continuous functions on an interval, [ilmath]\mathcal{C}[a,b][/ilmath] and the distance function:
    • [ilmath]d(f,g)=\sqrt{\int^b_a\vert f(x)-g(x)\vert dx}[/ilmath] for [ilmath]f,g\in\mathcal{C}[a,b][/ilmath]
    • Let [ilmath]a=-1[/ilmath] and [ilmath]b=1[/ilmath] (WLOG)
    • We can then see that the sequence of functions [ilmath](f_n)_{n=1}^\infty[/ilmath] where each [ilmath]f_n:[-1,1]\rightarrow[0,1]\subset\mathbb{R} [/ilmath] given by:
      [ilmath]f_n(x)=\left\{\begin{array}{lr}0 & \text{for }x\in[-1,0] \\ nx &\text{for }x\in(0,\frac{1}{n}] \\ 1 & \text{otherwise}\end{array}\right.[/ilmath]
      • Has a limit (note that: [ilmath]\lim_{n\rightarrow\infty}(f_n)=f[/ilmath] with [ilmath]f(x)=\left\{\begin{array}{lr} 0 & \text{for }x\in[-1,0] \\ 1 & \text{otherwise}\end{array}\right.[/ilmath] and that this [ilmath]f[/ilmath] isn't continuous (in [ilmath](\mathbb{R},\vert\cdot\vert)[/ilmath]) anyway!)
      • and that limit, [ilmath]f[/ilmath] isn't continuous, this we have shown that [ilmath]\mathcal{C}[-1,1][/ilmath] isn't complete. (and by translation/scaling as needed, [ilmath]\mathcal{C}[a,b][/ilmath] isn't complete)


  1. Functional Analysis - George Bachman and Lawrence Narici
  2. Analysis - Part I: Elements - Krzysztof Maurin