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A semi-metric space (M,d) satisfies all of the conditions of a metric space except it need NOT satisfy $d(f,g)=0 \iff f=g$.

Give 3 different examples of semi-metric spaces which are NOT metric spaces.

I understand the difference between a semi-metric space and metric space but I can not come up with 3 concrete examples.

Do the definitions of open, closed, dense and connected all still make sense in a semi-metric space without any alterations from the metric space setting?

I believe that the definition of open will still make sense in a semi-metric space. But the definition of closed, dense, and connected will not.

Am I right or wrong?

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  • $\begingroup$ Think of function spaces $\endgroup$
    – Matthew Levy
    Oct 13, 2014 at 1:54
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    $\begingroup$ Other which is pretty simple. $\forall x,y\in M. \,d(x,y)=0$ $\endgroup$
    – Jose Antonio
    Oct 13, 2014 at 2:22
  • $\begingroup$ How would I proove that these properties are true in a semi-metric space $\endgroup$
    – Username Unknown
    Oct 13, 2014 at 3:03

3 Answers 3

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In $\Bbb R$ ; $d(x,y)=|\sin x-\sin y|$.

Or $d(x,y)=|f(x)-f(y)|$ where $f$ is not injective. ($f(x)=x^2$, $f(x)=\cos x$...)

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  • $\begingroup$ Perfect. I don't want to over exhausted the trig functions what would be another example $\endgroup$
    – Username Unknown
    Oct 13, 2014 at 1:55
  • $\begingroup$ See the edit... $\endgroup$
    – Hamou
    Oct 13, 2014 at 1:56
  • $\begingroup$ Perfect. My question now becomes if my claim for the second part of the question is true $\endgroup$
    – Username Unknown
    Oct 13, 2014 at 1:58
  • $\begingroup$ A semi-metric define a Topology, hence those definition. $\endgroup$
    – Hamou
    Oct 13, 2014 at 2:02
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For any $p \geq 1$, $d(f,g) = (\int |f(x) - g(x)|^p dx)^{1/p}$ defines such a metric.

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$$d(f,g) = |f(x_0)-g(x_0)|$$ for some fixed point $x_0$

$$d(f,g) = \int_D|f(x)-g(x)|d\mu$$ for integrable functions f,g $$d(A,B) = \inf \{ \tilde d(x,y):\forall x\in A \text{ and } \forall y\in B\}$$ for some metric $\tilde d$

Not 100% sure about the last one..

All strictly topological ideas will not change, but the defined open sets of the topology in general will change for a semi-metric space. I believe semi-metric spaces still satisfy certain separation axioms.

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