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Topology: An Inquiry-Based Approach

Steven Schlicker

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Section Introduction

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Metric spaces are particular examples of topological spaces. A metric space is a space that has a metric defined on it. A metric is a function that measures the distance between points in a metric space.
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We are familiar with one special metric, the Euclidean metric dE in R2 where
dE((x1,x2),(y1,y2))=(x1−y1)2+(x2−y2)2.
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Figure 3.1. The Euclidean distance between (x1,x2) and (y1,y2) and the Euclidean unit circle in R2.
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Using this metric, the distance between two points (x1,x2) and (y1,y2) is the length of the segment connecting the points, while the unit circle (the set of points a distance 1 from the origin) looks like what we think of as a circle as illustrated in Figure 3.1.
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As we will see, there are many other metrics that can be defined on Rn, or on other sets.
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Preview Activity 3.1.

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Consider the function dT that assigns to each pair of points in R2 the real number
dT((x1,x2),(y1,y2))=|x1−y1|+|x2−y2|.
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This function dT is sometimes called the taxicab metric or distance because the distance between points x and y can be thought of as obtained by driving around a city block rather than going directly from point x to point y.
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Any distance function should satisfy certain properties: the distance between two points should never be negative, the distance from point A to point B should be the same as the distance from point B to point A, the shortest distance between two points A and B should never be more than the distance from A to some point C plus the distance from C to B, and the distance between points should only be zero if the points are the same. In this activity, we determine if dT has these properties. Let x=(x1,x2) and y=(y1,y2) in R2.
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(b)

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Prove that dT(x,y)=dT(y,x).
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(c)

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Prove that dT(x,y)=0 if and only if x=y.
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(d)

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Let z=(z1,z2) in R2. Read the proof of Lemma 3.2 (below) and then use Lemma 3.2 to show that
dT(x,y)≤dT(x,z)+dT(z,y).
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(Do you have any questions about the proof of the lemma?)
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Proof.
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Let a and b be real numbers. To prove the lemma we consider cases.
Case 1: a≥0 and b≥0
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In this case a+b is nonnegative and so |a|=a, |b|=b, and |a+b|=a+b. Then
|a+b|=a+b=|a|+|b|.
Case 2: a≤0 and b≤0
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In this case a=−a′ and b=−b′ where a′ and b′ are nonnegative. It follows from Case 1 that
|a+b|=|−(a′+b′)|=|a′+b′|=a′+b′=|a′|+|b′|=|−a′|+|−b′|=|a|+|b|.
Case 3: One of a or b is positive and the other negative
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Without loss of generality we assume a>0 and b<0. Again we consider cases. Note that b<0 implies a+b<a.
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    Suppose b≥−2a. Then a+b≥−a and so −a≤a+b<a. It follows that
    |a+b|≤a=|a|<|a|+|b|.
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    The last case is when b<−2a. In this case −b>2a and so
    |b|=−b>2a=2|a|>|a|.
    Then a+b<a=|a|<|b|. Finally, a>0 implies a+b>b=−|b|. So
    −|b|<a+b<|b|
    and
    |a+b|≤|b|<|a|+|b|.
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This proves our lemma for every possible pair a, b.
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(e)

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A picture to illustrate the taxicab distance dT between (points x1,x2) and (y1,y2) is shown in Figure 3.3. Draw a picture of the unit circle (the set of points a distance 1 from the origin) using the Taxicab metric. Explain your reasoning.
Figure 3.3. The taxicab distance between (x1,x2) and (y1,y2) in R2.
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The taxicab metric can be extended to Rn for any n≥1 as follows. If x=(x1,x2,…,xn) and y=(y1,y2,…,yn) are in Rn, then the taxicab distance dT(x,y) from x to y is defined as
dT(x,y)=|x1−y1|+|x2−y2|+⋯+|xn−yn|=∑i=1n|xi−yi|.
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