To be homeomorphic or isometric means "to be the same" ? What about $L^3$ and $L^{3/2}$ or $(\mathbb R^2,\|.\|_1)$ and $(\mathbb R^2,\|\mathbb \|_2)$?

Question

I'm confuse with something : my topological teacher told us that *To be homeomorphic means to be the same". Does it also hold for normed vector spaces or to be "the same" they must be isometric ? I really have problem with this because normed space are topological spaces. Let few example to illustrate :

1) Let $(\mathbb R^2, \|\cdot \|_1)$ and $(\mathbb R^2, \|\cdot \|_2)$ where $\|(x,y)\|_1=|x|+|y|$ and $\|(x,y)\|_2=\sqrt{x^2+y^2}$. I know that both norms are equivalent, and thus $\mathcal T_{1}=\mathcal T_2$ where $\mathcal T_i$ is the topology of $\mathbb R^2$ induced by $\|\cdot \|_i$. Therefore, $X_1:=(\mathbb R^2,\mathcal T_1)$ and $X_2:=(\mathbb R^2,\mathcal T_2)$ are homeomorphic, and since $\mathcal T_1=\mathcal T_2$, they are in fact the same, so $X_1=X_2$ makes sense. But they are not isometric, so there are not the same ?

2) Now, Riesz-representation says for example that $(L^1)'$ and $L^\infty $ (where $(L^1)'$ means the topological dual of $L^1$) are isometric, and thus $(L^1)'=L^\infty $ in the sense that there are the same.

3) Now, I know that for example $L^3$ and $(L^{3})''$ are isometric (and thus are the same). I know that $L^3\hookrightarrow (L^3)'$ and thus $L^3$ and $(L^3)'=L^{3/2}$ are homeomorphic. Now it doesn't make any sense to say that $L^3$ and $L^{3/2}$ are the same since for example there are function that are in $L^3$ that are not in $L^{3/2}$ and reciprocally.

4) The thing is I'm not sure that the bijection between $L^3$ and $(L^3)'$ is an homeomorphism. If it's not, suppose that $(X,\|\cdot \|_X)$ and $(Y,\|\cdot \|_Y)$ are homeomorphic but not isometric. Can we say that there are the same or not ? If not, what is the subtelty in the example 1) ?

Could someone explain me the difference subtlety in those 4 examples ?

Answer 1

"Sameness" in mathematics is almost always subject to a certain category of objects, such as topological spaces or groups. In the beginning, when algebraic structures such as vector spaces or groups were identified with other vector spaces and groups, the word used to express "sameness" was "isomorphic" - a nice Greek sounding word. For all algebraic purposes, two isomorphic groups are the same. Then came along topology, with its definition of topological spaces, and since the word "isomorphic" was already spoken for by algebraists, they used the word "homeomorphic" to describe sameness of topological spaces. The word "isometric" is yet another word to describe sameness, but this one is usually used for metric spaces, which are special topological spaces which also contain a notion of distance. On top of that, in the context of normed spaces, the word "isometric" means that the unit ball with respect to one norm is a linear transformation of the unit ball of the other norm. Thus, $(\mathbb{R}^2,\|\cdot\|_1)$ is homeomorphic to $(\mathbb{R}^2,\|\cdot\|_2)$ as topological spaces, because it is possible to write down a continuous-with-continuous-inverse function that maps the unit ball of one norm onto the other, but they are not isometric as normed spaces, because there is no linear transformation that carries the unit ball of the $\|\cdot\|_1$ norm to that of the $\|\cdot\|_2$ norm. This is because linear transformations carry polygons to polygons.

So one must always be mindful of the context. When people say "The Banach space $X$ is isomorphic to the Banach space $Y$", they mean that not only as vector spaces the spaces are isomorphic, but also as normed spaces they are "the same" - meaning that their norms are equivalent in a sense that can be made precise. Even so, it is important to keep in mind that equivalence of norms only guarantees that the topologies induced by the norms are the same, not the metrics induced by the norms. This is unfortunate as it is misleading. Unfortunate, because what is usually more interesting about a normed space is the metric induced by its norm, rather than the mere topology. To illustrate this point, we only need to remind ourselves that all finite dimensional real normed spaces are isomorphic, which means that their unit balls are homeomorphic, but as was already mentioned above, they are not isometric as metric spaces. Misleading, because it has caused generations of mathematicians to read into the notion of "isomorphic Banach spaces" much more than there is to it, by attaching to the theory the word "geometric". Thus one finds numerous titles that include the phrase "the geometry of Banach spaces" whereas in fact virtually all the results mentioned are about the topology of Banach spaces. Topology is not geometry. For geometry we need a metric. Riemann knew that, and Einstein knew that. To deduce from equivalence of norms anything about the geometry is actually impossible, by definition.

Answer 2

To be homeomorphic is to be the same as topological spaces: If $X$ and $Y$ are homeomorphic we cannot distinguish them by their topologies or topological properties, like connectedness, compactness etc. E.g. if both are linear spaces, a homeomorphism need not even preserve any linear structure etc., it just needs to respect the topology.

If normed spaces are (linearly) isometric, this is much stronger: it means that we cannot distinguish the spaces by properties definable by the norm and its distance. So e.g. is one is complete so is the other, if one has a smooth norm, so has the other etc. It implies that they're also homeomorphic but it's much stronger. Your first example is an example of non-isometric but homeomorphic spaces.

Most proporties in functional analysis are concerned with properties connected to the norm, metric and linear structures, not only the topology. In fact if we look at the topology of linear spaces, a well-known theorem tells us that all completely metrisable, separable locally convex topological vector spaces are homeomorphic. But $\ell^p$ is not isometric to $\ell^q$ for $p \neq q$. So with isometry we "see" more different spaces, for a suitable definition of different.

Answer 3

To be homeomorphic is to be the same as far as topology goes. So if you have two topological spaces $X$ and $Y$ which are homeomorphic, then whenever $X$ has some topological property $p$, then so does $Y$ (and vice versa).

This says nothing about non-topological properties of those objects. For example, $\mathbf C$ and $\mathbf R^2$ are clearly homeomorphic (and even isomorphic as abelian groups), but as rings, they are very different.

This phenomenon is not restricted to topology. The general notion is that of an isomorphism. In very abstract terms, if you have two mathematical objects $A,B$ in a "scope" you have "maps" $f_1,f_2$ going both ways, such that the compositions $f_1\circ f_2$ and $f_2\circ f_1$ are "identity" on the respective spaces and $f_1$ and $f_2$ both "preserve" the properties the "scope" knows, then $A,B$ are effectively the same (because anything we prove about $A$ can be translated to $B$ and vice versa).

Some elementary examples are group isomorphisms, ring isomorphisms, homeomorphisms (for topological spaces), isometries (for metric spaces) and linear isomorphisms (for vector spaces). For more complex objects, like Banach spaces, you have more complicated notions like "topological isomorphism" (in this context, a linear isomorphism which is also a homeomorphism) or "isometric isomorphism". There are also more vague notions of isomorphism like a "quasi-isometry" between metric spaces, which may not even be a bijection, but is still a notion of isomorphism.

The vague description I gave in the third paragraph can be made more formal by the means of model theory or category theory. The model-theoretic approach is more intrinsic, but a bit clumsy for more complicated structures (typically, it is used for purely algebraic structures), the category-theoretical formalisation is extrinsic, but much more flexible (it is used for all kinds of structures). Both of these are a bit too complicated to explain in this post.