Suppose $X$ and $Y$ are two Banach spaces. If there is an isometric isomorphism, then we can say that these two spaces are same. Is there any other condition which will say that these two spaces are same?

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    $\begingroup$ Think about what you mean by "same". $\endgroup$ – David Mitra Feb 28 '14 at 12:53
  • $\begingroup$ if there is isomorphic isometry, we write $X=Y$. is there any other sense to write $X=Y$? $\endgroup$ – partha Feb 28 '14 at 13:05
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    $\begingroup$ Analogous question: when are two groups equal? Answer in both cases: the underlying sets are equal and the extra structure coincides. $\endgroup$ – Rasmus Feb 28 '14 at 13:07
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    $\begingroup$ $(L^p)^*= L^q$ but the underlying sets are not same. $\endgroup$ – partha Feb 28 '14 at 13:10

I think there is something to be fixed with the notions of "equality or the same " and "essential equality or isomorph to" and "isomorphic and isometric to"

1) equality is a "defined" as an axiom ( extension axiom, X=Y iff they have the same elements i.e. $X\supset Y, X \subset Y$)

2) Isomorphic equality between two Banach space means there exists an homeomorphism between the two spaces. So are -essentially- preserving the topology of the space BUT you don't care of the norms over them.

3) Isometric isomorphism is like case 2) but you want to preserve norms.

In fact there exists examples of Banach space which are the same in the sense of 2) but not 3) here for example

and an easy example of two banach spaces for whose 3) is valid but 1) not. (like your example).

When you talk about reflexivity, you are considering a special case of 3) (so maybe it can be a 4) case :) - you want that 3) is true for a particular function, the canonical immersion. )

I want to stress the fact that 1) is purely set-theoretic. Two sets are the same if they are the same, not under a "change of name".

Maybe if we talk about couples it can be more clear. In 2) we are preserving the couple $(B, \tau)$ where $\tau$ denotes a topology over $B$. $(B,\tau_1)$ and $(B,\tau_2)$ are not the same topological space (it's a couple!), but it is the same set $B$.


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