Can "vacuum be brought" from outer space? Instead of creating vacuum in chambers on earth, why can't vacuum be brought from outer space in chambers? Outer space pressure ranges from $10^{-6}$torr to $10^{-17}$torr very very low. Is it possible at all?
 A: A lot of things are possible if you want to throw a lot of money at it.
Is it practical? No.
First, I would estimate that a rocket would be thousands of dollars. Low Earth orbit would get you the high range ($10^{-6}$ torr), high Earth orbit would possible get $10^{-9}$ torr. The next difficulty is to have the container returned to Earth (it needs to be shielded from the heat of re-entry) and retrieved from where ever it landed (several thousand more dollars).
Of course, the container needs a good seal since air will be trying to find its way in.
A good vacuum system (about \$2000) can reach $20\cdot 10^{-9}$ torr for a small chamber in about 30 minutes. I have one for the mass spec at work.
A: To make a long story short - Yes, it is possible, but you would have no practical reason to do it.
To make a long story long, read on.
While space is not a perfect vacuum, like you stated, it is close, and many applications of earth-generated vacuums are for simulating the conditions of outer space.
In theory, one could put a pressure vessel in a orbital - or even suborbital - trajectory, have it open to the "vacuum" of space, re-seal it, and then recover the vessel planetside.  However, this would be impractical for a number of reasons.
1) Planning, launching, recovering, etc. for a space mission takes months.  This is much longer than generating a vacuum with a pump takes.
2) Vacuum Chambers, due to the amount of strength needed to handle the large pressure differential, are very heavy.  Launching a rocket is not cheap, and the expenses increase with the increases in weight in the payload.
As far as specific costs go, I'm no expert, so I can only give very broad estimations.
Even if the cost of launching a rocket to obtain a vacuum or purchasing/installing a vacuum chamber, pumps, etc. are about the same (as I imagine they are), the vacuum chamber would be more practical because it can be used almost on-demand (no waiting for a rocket launch) could be reused multiple times (wheras a new rocket would have to be purchased and launched), and would not need to be transported from recovery site to the "site of use", like a launched pressure vessel.
A: Probably not. One problem is outgassing:
http://en.wikipedia.org/wiki/Outgassing#Outgassing_in_a_vacuum
Even if you built an airtight container, took it out into space, opened it to let the gas out, sealed it, and brought it back to Earth, you would still have the vacuum destroyed by outgassing.  Small atoms trapped in the material of the container itself would constantly be diffusing to the inner surface and ejected into the interior of the container.
To some extent this can be reduced by allowing the materials to outgass for a long time in a vacuum so that the impurities in the container slowly leave.  I don't know how long it takes for this to reach a steady state.
(There may be other more serious problems that an expert could tell you.  My knowledge about vacuum techniques is very modest.)
A: Never mind cost in dollars, it costs more in energy.  Assume you had perfectly efficient pumps.  To create a vacuum you have to push the air out of your chamber, if we assume we want $1 \text{ L}$ of vacuum, this takes 
$$ W = PV = (1 \text{ atm}) (1 \text{ L}) = 10^9 \text{ erg} $$
of work.
But if you want to get your vacuum from space, even assuming you had a perfectly efficient space elevator, you first have to lift your vessel into space, seal it, and bring it back.  On the way down you'll have to do work against the buoyant force of your vessel.  This will require work
$$ W \sim \rho V g H_0 = (1 \text{ kg/m}^3) ( 1 \text{ L} ) ( 9.8 \text{ m/s}^2 ) ( 9 \text{ km}) \sim 10^9 \text{ erg} $$
Where $H_0$ is the scale height of the atmosphere, that is, the exponential decay with height.  We can estimate the scale height by a simple statistical mechanical argument.  If the atmosphere were in equilibrium, all of its energy states should be populated according to the boltzmann factor $e^{E/kT}$, so the atmosphere's density should be, roughly,
$$ \rho(h) = \rho_0 e^{-m g h}{kT} = \rho_0 e^{-h/H_0} $$
where $mgh$ is the gravitational energy of a nitrogen molecule.  This gives
$$ H_0 = \frac{ kT}{mg } \sim 9 \text{ km} $$
The order of magnitude estimate comes out a wash (they are inherently the same calculation done two different ways), but they are a wash assuming we had a space elevator.  We don't have space elevators, we have rockets, which loose lots of energy to drag.  Rockets cost a lot more energy. So "space vacuum" should always come out behind "earth vacuum".
A: It's physically possible, but the chemical facility maintained a large LN2 tank with an appliance like oil diffusion pump to maintain the vacuum volume. so economically it is a bad idea.
