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$Gkmommwigknogkkon R Xr ->>> http://shurll.com/73ftj If you do not want the website to be able to write to certain folders, 555 would be a better setting, and you would use 444 on files that you do not want written by the web processuser group world r+w r 4+2+0 4+0+0 0+0+0 = 640 If no one has EXECUTE, nothing will changeYou don't actually help any further with hints on how to prove that the cardinality is the sameCHMOD also has the ability to be run as alphanumeric/symbolic mode$rn = frac1{sqrt{n^2+1}}$, and we define the following function: $$h(x)=begin{cases} r{2n} & exists n: x="rn" r{2n+1} & exists n: x="qn" x &text{otherwise}end{cases}$$ Now we can finally describe a list of bijections which, when composed, give us a bijection between $mathbb R$ and $mathbb{R^N}$chmod =rw will set READ (4) and WRITE (2) to the usual defaults–TonyK Jan 29 '13 at 21:09 CutieKrait: I was the one who downoted youGeorge Gates January 17th, 2013 at 11:56 I understand the three permissions together like r= read w="write" x="execute" (rwx)Now every number except of form $frac{m}{2^n}$ has a unique binary representation since those can't be represented in finite way(LogOut/Change) You are commenting using your Google+ accountup vote 4 down vote NOTE: This doesn't work First, map all the $mathbb R$s to $(0,1)backslash mathbb Q$then by axiom of choice, there are some $Asubseteqmathbb{R}$ and some bijection $g:Atomathbb{R^2}$First change binary expansion of both $x,y$ into the expansion which never ends in zeroes (if it needs)–TonyK Jan 29 '13 at 20:58 TonyK: Can you do $xmapsto(x,0)$ for the injection $(0,1)rightarrowtail(0,1)^2$, since $0notin (0,1)$? –user59083 Jan 12 '14 at 17:52 add a comment If your permissions before were 777, they would now become 776$mathbb{(0,1)to R}$ by any bijection of this sort, e.gThat last group setting can probably be made 0, as nothing other than the web process really should be hitting your files or folders–Asaf Karagila Aug 14 '14 at 11:56 Ohh I didn't notice the actual problem, I thought it was $mathbb{R}^{N}to mathbb{R}$ for some finite $N$, ok lets see if I can extend for $mathbb{R}^{mathbb{N}}$ –dragoboy Aug 14 '14 at 11:59 add a comment Now, we know that $mathbb{N^N}$ can be identified with the real numbers, in fact continued fractions form a bijection between the irrationals and $mathbb{N^N}$ For instance, $0.7170707070707070.$ and $0.7079797979797979.$ both unpack to $(0.777777.,0.1)$I think I have it right nowIf you wanted to restrict things to just your user, you could set 700 on your folders and 600 on your filesI will give a brief explanation: As you have described the permissions are usually displayed as nine letters in a combination of r w or x, there is sometimes a seventh letter at the beginning, if for example the first letter is d you are looking at a directory(d for directory) etcHOWEVER! Please can you re-publish with not conflated into a single dash please in order to make it searchable? i.eWhy? $$mathbb{(N^N)^Nsim N^{Ntimes N}sim N^N}$$ And the bijections above are easy to calculate (I will leave those to you, the first bijection is a simple Currying, and for the second you can use Cantor's pairing function)Not sure how to prove this thoughRepeat this process for the other user groups and combine we get (rwxrr == 744 : each group of three represents one number)So the usual approach is to show the existence of an injection in each direction, which guarantees the existence of a bijection by the SchroederBernstein theorem$mathbb{(0,1)setminus Qto (0,1)}$ by the decoding of $h$, i.eNow consider binary expansion$