Recall that an $F_{\sigma}$ set is, by definition, a countable union of closed sets.  We see that $\mathbb{Q}$ is an $F_{\sigma}$ set: $\mathbb{Q} = \cup_{q \in \mathbb{Q}} \{q\}$.  Is $\mathbb{R} \backslash \mathbb{Q}$ an $F_{\sigma}$ set? Note: $\mathbb{Q}$ and anything in $\mathbb{R} \backslash \mathbb{Q}$ are hollow.  I have no idea how to do this.  I know the goal is to use Baire Caegory Theorem.  I think that since anything in $\mathbb{R} \backslash \mathbb{Q}$ is hollow, we have that $\mathbb{R} \backslash \mathbb{Q}$ is a collection of hollow sets, $\{F_n\}$ in $\mathbb{R}$. I don't know how to show the sets are closed or the collection is countable.  Then by Baire Category Theorem, we know that $\cup_{n=1}^{\infty}$ is hollow, but $\mathbb{Q}$ being hollow means $\mathbb{R} \backslash \mathbb{Q}$ is dense  $\mathbb{Q}$ is hollow, that tells us that $\mathbb{R} \backslash \mathbb{Q}$ is dense (which seems to contradict the note, but maybe I'm just not reading it right).   If this isn't the best way to approach the problem, please let me know the better alternative.  EDIT: I think it might be best to prove it as such: Suppose $\mathbb{R} \backslash \mathbb{Q}$ is a $F_{\sigma}$ set.  Then $\mathbb{R} \backslash \mathbb{Q} \cup \mathbb{Q}$ is a countable collection of closed collow sets so by Baire Categoty Theorem, the union of this collection of sets is hollow, but $\mathbb{R} \backslash \mathbb{Q} \cup \mathbb{Q}$ is just $mathbb{R}$, which we know is dense, so we've reached a contradiction
I am trying to prove that the set of irrational numbers $\mathbb{R} \setminus \mathbb{Q}$ is not an $F_{\sigma}$ set. Here's my attempt:  Assume that indeed $\mathbb{R} \setminus \mathbb{Q}$ is an $F_{\sigma}$ set. Then we may write it as a countable union of closed subsets $C_i$: $$ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty}  \ C_i $$ But $\text{int} ( \mathbb{R} \setminus \mathbb{Q}) = \emptyset$, so in fact each $C_i$ has empty interior as well. But then each $C_i$ is nowhere dense, hence $ \mathbb{R} \setminus \mathbb{Q} = \bigcup_{i=1}^{\infty}  \ C_i$ is thin. But we know $\mathbb{R} \setminus \mathbb{Q}$ is thick, a contradiction.  This seems a bit too simple. I looked this up online, and although I haven't found the solution anywhere, many times there is a hint: Use Baire's Theorem. Have I skipped an important step I should explain further or is Baire's Theorem used implicitly in my proof? Or is my proof wrong? Thanks.  EDIT: Thin and thick might not be the most standard terms so:  Thin = meager = 1st Baire category
Indicator icons do not appear after upgrade to Ubuntu 17.10 If $E[X|Y]=Y$ almost surely and $E[Y|X]=X$ almost surely then $X=Y$ almost surely Assume that $X$ and $Y$ are two random variables such that $Y=E[X|Y]$ almost surely and $X= E[Y|X]$ almost surely. Prove that $X=Y$ almost surely.   The hint I was given is to evaluate: $$E[X-Y;X&gt;a,Y\leq a] + E[X-Y;X\leq a,Y\leq a]$$  which I can write as: $$\int_A(X-Y)dP +\int_B(X-Y)dP$$ where $A=\{X&gt;a, Y\leq a\}$ and $B=\{X\leq a,Y\leq a\}$.  But I need some more hints. How to compress / reduce the size of JPEG photos for archiving? I have 50,000 high resolution JPEG photos, where a couple of them might occasionally be needed, about once a year.  I wanted to zip them to save disk space, except that zipping gives no space benefit - so trying to reduce the images disk usage using winzip, winrar or 7zip was not successful.  Is there any software or algorithm similar to zip to compress image size on hard disk for storage without loosing any image information? Subsetting a 2D numpy array Is there a way for non-root processes to bind to "privileged" ports on Linux? It's very annoying to have this limitation on my development box, when there won't ever be any users other than me.  I'm aware of , but none of them do exactly what I want:    (The version in Debian testing, 1.0, only supports IPv4)  (the "nat" table is not yet implemented for ip6tables, the IPv6 version of iptables) sudo (Running as root is what I'm trying to avoid) SELinux (or similar). (This is just my dev box, I don't want to introduce a lot of extra complexity.)   Is there some simple sysctl variable to allow non-root processes to bind to "privileged" ports (ports less than 1024) on Linux, or am I just out of luck?  EDIT: In some cases, you can  to do this. I am trying to make a snowball that triggers a command when it hits a player in 1.15, but I don't know how to do that. When I do execute at @e[type=snowball] run gamemode spectator @a[distance=..2] it only runs the command if it goes right next to them. Derivatives of $ \frac{1}{r} $ and Dirac delta function I am trying to understand the formula \begin{equation} \nabla^2\left(\frac{1}{|{\bf r}-{\bf r}'|}\right) = - 4 \pi \delta(\bf{r}-\bf{r}'), \qquad\qquad {\rm (I)} \end{equation} where ${\bf r}=(x,y,z)$. This is something heavily used in electrostatics and the steps to 'show' this is often the following: The first derivative reads \begin{equation} \nabla \frac{1}{| {\bf r} - {\bf r}' |} = - \frac{ {\bf r} - {\bf r}'}{| {\bf r} - {\bf r}'|^3} \end{equation} And taking the second derivative gives zero, except for the singularity at ${\bf r} = {\bf r'}$. Then from the divergence theorem we have \begin{equation} \int dV \,  \nabla^2 \frac{1}{| {\bf r} - {\bf r'}|} = \int dS  \,\,{\bf n} \cdot ( \nabla\frac{1}{|{\bf r} - {\bf r}'|}) = -4 \pi \end{equation} where the integration is performed over a sphere centered at ${\bf r}'$. Q1: Is there a more direct proof for equation (I)? Then my main question is about the separate second-order differentials. For instance, we can obtain, by direct computation \begin{equation} \partial_x^2 \, \frac{1}{| {\bf r} - {\bf r}'|} = \frac{ 3 (x-x')^2 }{| {\bf r} - {\bf r'} |^5} - \frac{1}{| {\bf r} - {\bf r}'|^3} \end{equation} Q2: Should there be a $\delta$ function on the r.h.s of this equation? Inverting direction of mouse scroll wheel Is there a way to make a mouse scroll wheel invert its direction? I'd like to scroll upwards and have that action scroll downwards and vice-versa. There is no setting on the Mouse control panel that makes this possible.   Any pointers to a hack or a particular mouse model that has such a setting would be appreciated. I am using Windows 7. How can I create that glitter tears effect and the sparkles? $A \cup B$ and $A \cap B$ connected $\implies A$ and $B$ are connected Let $X$ be a topological space. Let $A$ and $B$ be two subsets of $X$ such that: $A$ and $B$ are closed, $A \cup B$ is connected, $A \cap B$ is connected     Prove that $A$ and $B$ are connected.   Thoughts:  Suppose that $A$ is disconnected. Since $A$ is closed, there are two non-empty, disjoint, closed sets $E$ and $F$ in $X$ such that $A = E \cup F$. Then I attempted many things but all were not really useful.  Please help. Thanks. Recommended books for advanced undergraduate electrodynamics this works in viewport but not in render.  import bpy from math import *  def frame_handler(scene):     z = scene.objects['Lift'].location[2]     scene.objects['Text'].data.body = str(floor(z / 4.0) - 2)  bpy.app.handlers.frame_change_pre.clear() bpy.app.handlers.frame_change_pre.append(frame_handler)   How to get it working in render? I've read  and changed from "bpy.data.objects" to "scene.objects" as already seen in the script. What am I missing? Blender v. 2.81.16  Edit. Tried that depsgraph thing mentioned in comments and changed to frame_change_post:  import bpy from bpy.app.handlers import persistent from math import *  @persistent  # maybe not relevant but doesn't harm anything? def frame_handler(scene, depsgraph):     z = scene.objects['Lift'].evaluated_get(depsgraph).location[2]     scene.objects['Text'].data.body = str(floor(z / 4.0) - 2)  bpy.app.handlers.frame_change_pre.clear() bpy.app.handlers.frame_change_post.clear() bpy.app.handlers.frame_change_post.append(frame_handler)   Now the Text doesn't show up at all in rendered (but works in viewport). What is $\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^n$? I'm asked to find $\displaystyle\lim_{n\to \infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^n$.  It seems that the sum converges.  I managed to prove that $\forall n,1&lt;\sum_{k=1}^n \left(\frac{k}{n}\right)^n&lt;e$  But I don't know how to prove convergence, nor find the value of the limit. E: The repository 'http//archive.canonical.com precise Release' is not signed Cannot set LD_LIBRARY_PATH in /etc/profile.d Website is under maintenance - how to restrict access? Why does the zoom sometimes stop at a point? How do I mount an `img` created with /bin/dd of a hard drive? How to make child process die after parent exits? Suppose I have a process which spawns exactly one child process. Now when the parent process exits for whatever reason (normally or abnormally, by kill, ^C, assert failure or anything else) I want the child process to die. How to do that correctly?    Some similar question on stackoverflow:   (asked earlier)  (asked later)      Some similar question on stackoverflow for Windows: Why is usable RAM less than total RAM? My girlfriend bought a laptop last week. It's a core 2 duo with 4 GB  We installed vista 64bit, and one of the first things we did was right click on "My computer" to see gthe properties. Immediately we noticed something strange about her RAM, the line said: Installed memory (RAM): 4,00 GB (3,68 GB usable)  I told her not to worry, thinking it must be something about the laptop hardware (considering her vista installation came from the same DVD as mine, and I never noticed anything like that on my 4 GB desktop). One hour ago, it got worse. We looked at Properties again, and it now says: Installed memory (RAM): 4,00 GB (2,98 GB usable)  What does that mean? Are those 1,02 GB missing or being used by the system?  EDIT: There is a possibility that the sytem information is wrong. I just noticed that it reports an intel T6500 processor, when it's actually a T6400.  How can I find out how much RAM is really available to the system?  EDIT2: Checking the resource monitors, it says 1003 MB are reserved for the hardware. Is that good or bad? Thanks