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Metric spaces

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Poirot

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Feb 15, 2012
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Are all metric spaces open sets? How can this be proved?
 

dwsmith

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Feb 1, 2012
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Are all metric spaces open sets? How can this be proved?
If $X$ is a metric space, then $X-\emptyset$ is open, then $X$ is closed. Also, $X-\emptyset$ is closed so $X$ is open. So the whole set is both open and closed.
 
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Poirot

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Feb 15, 2012
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how do you know X without the empty set is closed?
 

dwsmith

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how do you know X without the empty set is closed?
It is a metric space ie a Topological space. What are the requirements for a topological space?
 
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Poirot

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Feb 15, 2012
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I'm learning about metric spaces before I go on to topological spaces. I thought there would be a proof independent of that.
 

Plato

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Jan 27, 2012
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I'm learning about metric spaces before I go on to topological spaces. I thought there would be a proof independent of that.
The statement that a set $O$ is open means that if $x\in O$ then some ball $\mathcal{B}(x;\delta)\subseteq O$.
By definition that is true for $\forall x\in X$.
It is vacuously true for $\emptyset$.
So the whole space and the emptyset are open.
 
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Poirot

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Feb 15, 2012
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so a metric space is an open set X with a metric defined on it
 

Fantini

"Read Euler, read Euler." - Laplace
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Feb 29, 2012
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No, a metric space is an ordered pair $(X,d)$, which is a set $X$ with a function $d: X \times X \to \mathbb{R}$ such that:

$(i) \quad \; \; d(x,y) \geq 0;$
$(ii) \quad \; d(x,y) = 0 \iff x=y;$
$(iii) \quad d(x,y) = d(y,x);$
$(iv) \quad \, d(x,y) \leq d(x,w) + d(w,y),$

where $x,y,z \in X$.
 
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Poirot

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Feb 15, 2012
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Well, if the definition doesn't stipulate that X is open, I don't quite see how X is open 'by definition', as Plato claims.
 

Plato

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Jan 27, 2012
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Well, if the definition doesn't stipulate that X is open, I don't quite see how X is open 'by definition', as Plato claims.
You are mixing up definitions.
A metric space is a pair: a set and a metric defined on that set.
Open sets are then defined in terms of that metric.
Thus the space must be open by definition.

A set is open if each of its points is in a basic open ball which is a subset of the set.
Well every basic open ball is a subset of the whole space so the whole space is an open set.

So once again, here is how it works.
We start with a pair: $(X,d)$.
Define a basic open ball, $\mathbb{B}(x_0;r)=\{x: d(x_0,x)<r\},~r>0$.
Then define what it means to say a set is open.
 

Poirot

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Feb 15, 2012
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very clear thank you.