MT2002 Analysis

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Convergence in a metric space

Just as a convergent sequence in R can be thought of as a sequence of better and better approximtions to a limit, so a sequence of "points" in a metric space can approximate a limit here.


A sequence (xn) of points in a metric space (X, d) converges to a limit α if the real sequence (|d(xn, α)| converges to 0 in R.


If you insist on "back to basics" this reads:
Given ε > 0 there exists NN such that if n > N then we have d(xn, α) < ε.


  1. For R with its usual metric this is the same as before.

  2. In C with the metric d(z, w) = |z - w|, consider the sequence (z, z2, z3, ...) with |z| < 1.
    Then this sequence converges to 0 ∈ C.

    e.g. Take z = (1+i)/2 so that |z| = 1/√2

    The points lie on a spiral.

    Proof that the sequence converges

    Look at the real sequence (d(xn, 0)) = (|zn- 0|) = (|z|n)→ 0 since |z| < 1.

  3. Look at the sequence in R2 given in the following way.
    x1= 2√3, y1= 3 and then define the later terms by 2/xn+1= 1/xn+ 1/yn and yn+1= √(xn+1yn).
    This gives a sequence which evaluates numerically to:
    ( (3.4642, 3.0000), (3.2154, 3.1057), (3.1596, 3.1325), (3.1460, 3.1392), (3.1426, 3.1408), 3.1418, 3.1414), ...)

    This sequence is based on the method used by Archimedes to calculate π.
    Start with x1= the semiperimeter of the "outside hexagon"
    Start with y1= the semiperimeter of the "inside hexagon"
    and then double the number of sides to get a 12-gon , a 24-gon, etc.
    Archimedes took the calculation up to n = 5 (corresponding to a 96-gon).

    In fact the sequence in R2 converges to the point (π, π).

This last result suggests the following.


Convergence in R2 with its usual metric d2 is "componentwise".

That is ((x1 , y1), (x2 , y2), ... )→ (α , β) if and only if
(x1 , x2 , ... )→ α and (y1 , y2 , ... )→ β.

( ⇒ ) Given ε > 0 we know that we have N so that if n > N then √(|xn- α|2+ |yn- β|2) < ε. But then we must have |xn- α| < ε and so (xn)→ α. Similarly for the other component.

Conversely: Given ε > 0 choose N so that if n > N then |xn- α| < ε and |yn- β| < ε. But then √(|xn- α|2+ |yn- β|2) < √(ε2+ ε2) = √2 ε and so we may make this as small as we like.

In fact this last result holds for any finite-dimensional space Rn and also holds for such spaces with any of the metrics dp. The situation for infinite-dimensional spaces of sequences or functions is different as we will see in the next section.

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JOC September 2001