Course MT4522 Metric and Topological Spaces

Solution 2

  1. In R2 you can simplify things by taking u = x - y and v = y - z with u + v = x - z.
    Then summing the coordinates of u and v and using |a + b| lte |a| + |b| in R gives the triangle inequality in d1.

    For d2 : |u + v|2 = |u|2 + |v|2 + 2 u.v and since u.v = cos(theta)|u||v| and cos(theta) lte 1 this is lte |u + v|2 and the result follows.

    For dinfinity : the maximum of the first coordinates of u+ v lte maximum of |coordinates of u| + maximum of |coordinates of v| and similarly for the second coordinates.

  2.   

    On the interval (0, 1) if |x - y| < 1/2 then d(x, y) = |x - y| otherwise it is 1 - |x - y| and so this is the same as the distance measured on a circle of perimeter 1.

    The same formula does not define a metric on the closed interval [0, 1] since d(0, 1) = 0 but these two points are not the same.

  3. d(r, s) = 1/ (largest power dividing |r - s|). Use (r - s) + (s - t) = (r - t) to prove the triangle inequality. All the other properties follow easily.

    d(0, pk) = 1/pk and this rarrow 0 as k rarrow infinity. That is, the sequence (p, p2, p3, ...) rarrow 0.

  4. Look at the picture:

    A point on the set d1 = 1 is inside the set d2 = 1 and so d2 lte d1 . Similarly (since the circle is inside the square) we have dinfinity lte d2 .


    Alternatively, look at a different picture:

    d2(P, Q) = c gte a + b = d1(P, Q).
    Also, both a lte c and b lte c and so
    dinfinity(P, Q) = max{a, b} lte c.


    Without a picture, (a + b)2 = a2 + 2ab + b2 gte a2 + b2 implies a + b gte sqrt(a2 + b2) and so d1gte d2 .
    Also if a gte b then max{a, b} = a and since a2 + b2 gte a2 we have sqrt(a2 + b2) gte a and so d2gte dinfinity .


    If (an) rarrow v in d1 then given epsilon > 0, thereexists N such that n > N implies d1(an ,v) < epsilon implies (since d1gte d2 and dinfinity) and so convergence also takes place in d2 and dinfinity .
    One can draw pictures to show that d1 lte sqrt2 d2 and d2 lte sqrt2 dinfinity .
    (One can of course argue without using pictures too!)


    Then if (an) rarrow alpha in dinfinity we have:
    (forall epsilon > 0) (thereexists n belongs N) ( n > N) (dinfinity(an, alpha) < epsilon) implies d2(an, alpha) < epsilon/sqrt2 and d1(an, alpha) < epsilon/2 and so convergence also takes place in d1 and d2 .
    Similarly, if convergence takes place in d2 it also takes place in d1 .
    Combining this with the earlier result, sequences convergent in one of these norms are convergent in all of them.

    1. The sequence (0.49, 0.499, 0.4999, ...) rarrow 0.5 and applying f gives the sequence (0.9, 0.9, 0.99, ...) rarrow 1 while f(0.5) = 1.
      Hence f is not continuous and indeed is discontinuous at any terminating decimal (but continuous elsewhere).
    2. Applying g to the same sequence gives (0.50111... , 0.500111... , 0.5000111... , ...) rarrow 0.5
      However, g(0.5) = 0.6111... = 11/18 which is different.
    3. Write 1.0 as 0.999... to see that this is actually the map h(x) = 1 - x and so it is continuous.