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I am currently working through chapter two of Principles of Mathematical Analysis (ed. 3) by Walter Rudin. My question comes from pages 30-31.

I know that a metric space must satisfy the definition:

Definition: A metric space is an ordered pair, $(X,d)$, where $X$ is a set (whose elements are called points), and $d$ is a metric. The metric must satisfy the following, given $x,y,z \in X$:

(a): $d(x,y) \ge 0$

(b): $d(x,y) = 0$ iff $x=y$

(c): $d(x,y) =d(y,x)$

(d): $d(x,z) \le d(x,y) + d(y,z)$

Rudin then provides the example that euclidean spaces, $\mathbb{R}^k$ are also metric spaces.

Following, he then specifies that any subset of a metric space is also a metric space.

After proving this statement, I then decided to come up with examples, and as an exercise, try to challenge myself to come up with counter-examples. One attempt that I made was: $\{ \boldsymbol{x} \} \subset \mathbb{R}^k$ where $\boldsymbol{x}$ is a $k$-tuple. This provides the conjecture:

Conjecture: Given any $\boldsymbol{x} \in \mathbb{R}^k$, the set $\{ \boldsymbol{x} \} \subset \mathbb{R}^k$ is a metric space.

Proof: $d(x,x) = |x - x| = |0| = 0$. Therefore, the only possible metric is $0$ and hence all the definition's conditions are satisfied.

I am uneasy about the proof I have constructed because in the definition, two (and once three) elements are drawn from the metric space. So in order for the above to hold, I essentially need to give this element several times.

So the primary substance of my question is: in a metric space, can we let an element "do two jobs at once?"

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3  
Yes, since nowhere in the definitions does it say "for distinct elements $x,y$ it holds that blah blah", or "for distinct three elements $x,y,z$ it holds that blih blih". And just for the fun of the extreme, the empty set is also a metric space. –  Ittay Weiss Jun 4 '13 at 4:16
    
@IttayWeiss I am making google searches to help explain why the empty-set is a metric space. I find several sources just "assume it away." Intuitively, it makes sense, since any metric will be true for the empty set since there is no elements to worry about. Could you provide the precise argument for the case? –  Jordan Mahar Jun 4 '13 at 4:40
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I think it'll be helpful if you read about vacuous truth. –  Vectk Jun 4 '13 at 4:57
    
@JordanMahar try to prove that the empty set is not a metric space. How can an axiom fail? –  Ittay Weiss Jun 4 '13 at 7:35

2 Answers 2

up vote 4 down vote accepted

You can always let an element do two jobs at once in this sense unless you’re checking a statement that explicitly rules out the possibility, and here you are not. Here, for instance, you have to check a requirement that $d(x,z)\le d(x,y)+d(y,z)$ whenever $x,y,z\in X$; there is no implication here that $x,y$, and $z$ must be distinct points. Indeed, the triangle inequality must hold whether they are distinct of not. The same goes for the other clauses of the definition.

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This would be another examples in which you can see @Brian's remarks are ruling again. Let $X\ne\emptyset$ and $$ d(a,b) = \left\{ \begin{array}{ll} 0 & \quad a=b \\ 1 & \quad a\ne b \end{array} \right.~~~~~~\text{or}~~~~~~d(a,b) = \left\{ \begin{array}{ll} 2 & \quad a\neq b \\ 0 & \quad a= b \end{array} \right. $$

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Nice example! +1 –  amWhy Jun 5 '13 at 0:11

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