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I found this picture

I found this picture when looking up topological spaces.

Is this picture actually supposed to be interpreted as decreasing sets? That is, all inner product spaces are normed vector spaces, all metric spaces are topological spaces etc?

But this would mean that every inner product space and normed vector space was a metric space. However, I don't recall inner product spaces and normed vector spaces having a metric, so it's not a metric space. Right?

On the other hand Inner product spaces do have norms, so it is a normed space. So it does make sense that it's a subset of Normed vector space.

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    $\begingroup$ In a normed vector space $(V,\|\cdot\|)$ the metric is $d(x,y)=\|x-y\|.$ $\endgroup$ – Thomas Andrews Apr 2 at 15:28
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    $\begingroup$ Right picture but none of these is a set. $\endgroup$ – Moishe Kohan Apr 2 at 15:29
  • $\begingroup$ Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/… $\endgroup$ – Moishe Kohan Apr 2 at 18:25
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If a vector space has an inner product, it has corresponding norm ($\|x\|=\sqrt{\langle x,x\rangle}$). Not all normed spaces are of this form (hence the proper inclusion)

If a vector space has a norm, it has a corresponding metric $d(x,y)=\|x-y\|$. Not all metric spaces are of this form (e.g. the discrete metric on $\mathbb{R}$ is not, and moreover not all metric spaces have a natural underlying vector space structure that is needed for a norm).

If a set has a metric, it has a topology (the smallest one that makes all balls $B(x,r)$ open sets), but many topological spaces do not come from any metric (though there are theorems that tell us exactly when this is the case).

So the "inclusions" tell us that some structures naturally give rise to another more widely applicable structure.

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If $(V,\lVert\cdot\rVert)$ is a normed space, then $V$ becomes a amtric space if we consider the distance $d$ defined as $d(v,w)=\lVert v-w\rVert$.

And not all norms come from inner products. Consider, for instance, in $\mathbb R^n$ the norm$$\bigl\lVert(a_1,\ldots,a_n)\bigr\rVert_1=\lvert a_1\rvert+\lvert a_2\rvert+\cdots+\lvert a_n\rvert.$$

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  • $\begingroup$ Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product? $\endgroup$ – Qwertford Apr 2 at 15:36
  • $\begingroup$ How do you defined an inner product starting frome the discrete metric? $\endgroup$ – José Carlos Santos Apr 2 at 15:37

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