Are $C[0,1]$ and $C[0,1)$ isomorphic?

Question

Consider the two vector spaces $C[0,1]$ and $C[0,1)$, the spaces of continuous functions on $[0,1]$ and $[0,1)$ respectively. Note I do NOT give either of these a topology, I am purely interested in their properties as vector spaces. My question is whether they are isomorphic? The motivation comes from considering $C[0,\infty)$ which is isomorphic to $C[0,1)$. It is very easy to show that $C[a,b]$and $C[c,d]$ are isomorphic for real constants $b>a$, $d>c$ but this got me thinking about this more general case.

Possible further questions would be:

What about $C(0,1)$ (open interval), is this isomorphic to either of the above?

If I consider the space of continuous functions on $[0,\infty)$ that converge to a finite limit at $\infty$ then this is isomorphic to $C[0,1]$ (either compactify the half line or use the usual arctan function to map to $C[0,1)$ and take limits at $1$). But what about the space of bounded continuous functions on the half line?

These cases seem to be easier to look at if we do include topologies but it's interesting that as far as I can tell, it becomes harder without.

See for instance How to show two infinite-dimensional vector spaces are not isomorphic for details. — Alex Vong, Oct 20 '17 at 07:07
I see - so if I can prove they have the same dimension (axiom of choice says this makes sense) I simply take a bijection between their bases. Is there a way to construct such a bijection in these cases explicitly? — Mathmo, Oct 20 '17 at 07:16
True, but is there an isomorphism that doesn't use bases given by the axiom of choice? — Mathmo, Oct 20 '17 at 07:56
Again, explicit is the problem. You can add the [set-theory] tag to attract the relevant experts. — Martín-Blas Pérez Pinilla, Oct 20 '17 at 08:19

Asaf Karagila · Accepted Answer · 2017-10-20T11:43:18.527

7

The answer is negative without the axiom of choice, and in particular there is no explicit way to define an isomorphism between the various spaces.

It follows from assumptions such as "Every set of reals has the Baire property" (which is consistent without the axiom of choice) that a vector space can have at most one Fréchet topology. This is a consequence of automatic continuity, which gives us that every linear operator between Fréchet spaces is continuous.

If $C(0,1)$ or $C[0,1)$ were isomorphic as vector spaces to $C[0,1]$, this vector space would have several distinct Fréchet topologies, which is a contradiction.

edited Oct 20 '17 at 11:43

answered Oct 20 '17 at 10:47

Asaf Karagila

405,794

This is one of the most interesting side effects of "every set of reals has the Baire property", that topology seemingly sticks its head into function spaces. Even when we're not inviting it. – Asaf Karagila Oct 20 '17 at 10:52
You are probably right, Andrés. – Asaf Karagila Oct 20 '17 at 11:43
Fascinating! One quick question before I accept the answer. We know AC implies positive result and the above shows a negative result is consistent (if we don't assume AC). Suppose we assume that $C[0,1)$ and $C[0,1]$ are isomorphic. Does this imply AC in ZF? – Mathmo Oct 21 '17 at 17:23
@Mathmo: No. Of course not. No "localized assumption" would ever imply AC. You can have that all the sets that you can ever dream of are well-orderable, but the axiom of choice fails very very very badly somewhere above that in the universe of sets. – Asaf Karagila Oct 21 '17 at 17:28
Yes that makes sense. Thanks! – Mathmo Oct 21 '17 at 18:38

Are $C[0,1]$ and $C[0,1)$ isomorphic?

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