# Metric and also Topological frameworks generated by a standard

While confirming that some normed rooms were full, 2 inquiries involved my mind. They connect the topological and also the statistics frameworks generated by a standard.

1) Is it feasible to locate 2 equal standards $\|\cdot\|_1$ and also $\|\cdot\|_2$ on a vector room $V \$ such that $(V \ ,\|\cdot\|_1)$ is full and also $(V \ ,\|\cdot\|_2)$ is not?

2) Is there a vector room $V \$ and also 2 non - equal standards such that $V \$ is full about both?

Below I'm thinking $V \$ a vector room over a subfield of $\mathbb{C}$. Additionally I recognize that the solution is no if we just take into consideration limited - dimensional vector rooms.

[Edit: I'm taking into consideration 2 standards equal if they specify the very same geography. I assume it is the common idea Jonas referred in his comment. ]

0
2019-05-18 21:23:56
Source Share

1) No. It is uncomplicated to show that equal standards generate both the very same convergent series and also the very same Cauchy series. (Written prior to Rasmus is solution was uploaded, yet uploaded later.)

2) Yes. One means to see this is to keep in mind that isomorphism courses of vector rooms depend just on straight measurement, so the inquiry totals up to locating 2 nonisomorphic Banach rooms of the very same straight measurement. There are great deals of instances of these. Every infinite dimensional separable Banach space has linear dimension $2^{\aleph_0}$. Nonetheless, as an example, $\ell^1$ and also $c_0$ are separable Banach rooms that are not isomorphic (as Banach rooms).

In fact, "total up to" had not been fairly exact. It is absolutely enough that the 2 Banach rooms are not isomorphic, yet it is not essential due to the fact that you are just asking that certain map (the identification in the initial solution) is not an isomorphism. So what I offered above is in fact more powerful. To simply address 2), you can simply take any kind of boundless dimensional Banach room and also generate a new standard using a boundless straight isomorphism with itself.

The solution over was thinking that equal standards are specified as in this PlanetMath article. If rather you suggested just that the rooms are homeomorphic in the standard geographies, as Jyotirmoy Bhattacharya believed, after that the instances mentioned over will not function. Nonetheless, there are additionally instances of sets of Banach rooms that have the very same straight measurement yet are not homeomorphic, and also this will certainly operate in either instance. As an example, $\ell^\infty$ and also $c_0$ are not homeomorphic due to the fact that $\ell^\infty$ is nonseparable. Both rooms have straight measurement $2^{\aleph_0}$. This was currently stated for $c_0$, and also for $\ell^\infty$ it adheres to due to the fact that $c_0$ installs in $l^\infty$ (which offers the lower bound on measurement) and also due to the fact that the cardinality of $\ell^\infty$ is $2^{\aleph_0}$ (which offers the upper bound).

(I'm currently rather sure this isn't what you desire, based upon your edit, yet this still offers an additional instance for the real inquiry along with a response to Jyotirmoy is comment.)

By the way, an additional means to see that $2^{\aleph_0}$ is a lower bound for the straight measurements of $\ell^1$ and also friends is to take into consideration the linearly independent set $\{(1,t,t^2,t^3,\ldots):0\lt t\lt 1\}$. Cardinality of the rooms offers an upper bound.

0
2019-05-21 03:48:27
Source