For $B\in \mathcal B(\mathbb R)$ (the borel $\sigma -$algebra), we denote $C_t(B):=\{\omega :[0,1]\to \mathbb R\mid \omega (t)\in B\}$. Let $$\mathcal F=\sigma \{C_t(B)\mid t\in [0,1], B\in \mathcal B(\mathbb R)\}.$$ Why $\mathcal A:=\{\omega :[0,1]\to \mathbb R\text{ continuous}\}$ do not belong to $\mathcal F$ ? I think that I have to prove that there are no countable collection of $\{B_i\}\subset \mathcal B(\mathbb R)$ and $\{t_i\}$ s.t. $$\mathcal A=\bigcup_{i=1}^\infty C_{t_i}(B_i),$$ but I don't know how to do it. I tried by contradiction, i.e. I suppose that $$\mathcal A=\bigcup_{i=1}^\infty C_{t_i}(B_i),$$ but I don't know what to do...
Consider $\mathbb R^{[0,1]}$ the space of all functions from $[0,1]$  to $\mathbb R$  and the cylindrical sigma algebra $\mathcal B$  on it. The question is: how to prove that $C[0,1]\notin \mathcal B$.
This question is in the spirit of , an example of an apparently non-deterministic system in Newtonian mechanics. Under certain restrictions, the  guarantees the existence and uniqueness of solutions to differential equations with initial conditions. However, Norton's dome does not satisfy these restrictions. The ball may sit on top of the dome for an arbitrary period of time before sliding down in a particular trajectory, without violating Newton's laws of motion.    Is there a plausible example of a system of interacting point-particles such that the equations of a point-particle are continuous but not , and therefore fail to have a unique solution? I accidentally closed it without voting, how can I find it? When I suggest declaring war on the target it says there's already a proposition to do so. How to check if Node connected to another node input in Python I am doing a process to some switch links in materials. I need to know if something is ever linked or not before to process. I can't find out how to :/ The script works well if something is linked only.  import bpy for m in bpy.context.object.material_slots:     mat = m.material     nodes = mat.node_tree.nodes      # get node input (socket)     alphasocket = mat.node_tree.nodes.get("Principled BSDF").inputs[18]     # get node link that connects into the socket     lnk = (link.from_socket for link in mat.node_tree.links if link.to_socket == alphasocket)      #add emission node     emi= nodes.new('ShaderNodeEmission')     emi.name= "emibak"     outepute= mat.node_tree.nodes.get("Material Output").inputs[0]      #make new link destination     #here 's where I cant find out how to check     if lnk != None:         print(mat.name + " alphanod found")        # mat.node_tree.links.new(lnk, emi.inputs[0])     else :         print(mat.name + " alphanod missing")      mat.node_tree.links.new(emi.outputs[0], outepute)     here is another way I am searching...  import bpy  for m in bpy.context.object.material_slots:      mat = m.material     nodes = mat.node_tree.nodes          print("**** Material " + mat.name+ " ****")      for mat_node in mat.node_tree.nodes:         if mat_node.type == 'BSDF_PRINCIPLED':             print("From " + mat_node.name )                         alphasocket = mat_node.inputs[18]             #print(alphasocket)             node_in = mat_node.inputs             #print(node_in)                    for n in alphasocket.links:                                print("Alpha connected to " + n.from_node.name)   This show me what is connected to the alpha channel but if nothing is connected I want to connect a new node. I lost my .keystore file? How to access new colour schemes in version 10? Splitting a semicolon-separated string to a dictionary, in Python I have a string that looks like this:  "Name1=Value1;Name2=Value2;Name3=Value3"   Is there a built-in class/function in Python that will take that string and construct a dictionary, as though I had done this:  dict = {     "Name1": "Value1",     "Name2": "Value2",     "Name3": "Value3" }   I have looked through the modules available but can't seem to find anything that matches.    Thanks, I do know how to make the relevant code myself, but since such smallish solutions are usually mine-fields waiting to happen (ie. someone writes: Name1='Value1=2';) etc. then I usually prefer some pre-tested function.  I'll do it myself then. I'm trying to get Intel and Nvidia graphics card working together with two monitors. One monitor should use the Intel card and the other the Nvidia card. Of course, I want to use only one keyboard and one mouse to switch between both screens. Bumblebee didn't work for me.  I don't want to use the xserver-xorg-video-all package because I need the Nvidia card for gaming and I can't use only one graphics card because I need one for KVM virtualisation and the other one for my Linux desktop  So I tried to install the drivers and to configure the Xserver manually. The Nvidia driver is working and I get a video output. But I don't get a video output with the Intel card.  I followed   The config files: Why is compactness in logic called compactness? In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set.   Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn't seem strong enough to justify the terminology.   Is there more to the choice of the terminology in logic than this analogy? Open-source alternative to Adobe InDesign There is some open-source software, similar to Adobe InDesign? It would be awesome if it do exist...  I know all the others like Inkscape, Gimp, etc... but for InDesign i have none.  Could you give me one direction? Convert a character digit to the corresponding integer in C Types of photon qubit encoding How many types of qubit encoding on photons exist nowadays? I know only two:   Encoding on polarization: $$ \lvert \Psi \rangle = \alpha \lvert H \rangle + \beta \lvert V \rangle $$ $$ \lvert H \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{H}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$ $$ \lvert V \rangle = \int_{-\infty}^{\infty} d\mathbf{k}\ f(\mathbf{k}) e^{-iw_k t} \hat{a}^\dagger_{V}(\mathbf{k}) \lvert 0 \rangle_\text{Vacuum} $$ Time-bin: $$ \lvert \Psi \rangle = \alpha \lvert 0 \rangle + \beta \lvert 1 \rangle $$ $$ \lvert 0 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$ $$ \lvert 1 \rangle = \int_{-\infty}^{\infty} dz\ f\left(\frac{t -z/c+\tau}{\delta t_{ph}}\right) e^{-i w_0 (t-z/c+\tau)} \hat{a}^\dagger(z) \lvert 0 \rangle_\text{Vacuum} $$   Is there anything else? When do we not put a comma before "because"? In what situation do you put a comma before "because"?   And, when do you not put a comma before "because"?   I've always assumed that you put a comma, because the clause introduced by "because" will always be independent, because it cannot stand on its own, because it wouldn't make sense on its own.   Is this the case? Could you think of an example where it doesn't take a comma? Anti-Matter Black Holes Assuming for a second that there were a pocket of anti matter somewhere sufficiently large to form all the type of object we can see forming from normal matter - then one of these objects would be a black hole.  Question is, would there be a difference between an anti matter black hole and a normal matter black hole - in terms of how would the matter/anti-matter make the black hole different, or would they be the same?  I would expect the answer to be that the black hole formed from anti matter would retain the anti-matter properties in such a way that if it was to merge with a black hole at same size formed from normal matter that they would annihilate  each other con convert into pure energy - Would that be a correct understanding? I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is $$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$. create multilingual node programmatically i'm trying to create a multilingual node using a php script.  I'm doing this:   $node = new stdClass();  $node-&gt;title = "Test ".mktime();  $node-&gt;type = "job";  $node-&gt;language = 'de'; // Or e.g. 'en' if locale is enabled  $node-&gt;uid = $user-&gt;uid;   $node-&gt;status = 1; //(1 or 0): published or not  $node-&gt;promote = 0; //(1 or 0): promoted to front page  $node-&gt;comment = 1; //2 = comments on, 1 = comments off  $node-&gt;path = array('alias' =&gt; 'das-ist-ein-test-'.mktime());  $node-&gt;body[$node-&gt;language][0]['value']   = "Body Body Body";  $node-&gt;body[$node-&gt;language] = text_summary("Body Body Body");  $node-&gt;body[$node-&gt;language]  = 'filtered_html';  node_save($node);   As long as i set $node->language to 'und' everything is working fine. But as soon as I set the language to 'de' or 'en' (both languages are activated in my Drupal installation) the body text doesn't get saved.  So my question is: can anybody provide me with a working example of how to store a multlingual node (de,en) into Drupal using program code.  Thank you Using \@-ified punctuation while using \frenchspacing After reading  about spacing and punctuation, I was left wondering about an issue that didn't seem to be addressed: using \frenchspacing together with the modified punctuation \@., \@?, etc.  As far as I can tell visually, using \frenchspacing together with these commands works correctly (i.e., it seems like nothing breaks, and the extra space is suppressed as normal), but are there problems that can arise from doing so?   Moreover, when using \frenchspacing, "should" I still use \@. and its friends where it would ordinarily be necessary, in case someone wants to remove \frenchspacing later? Would accommodating (hypothetical) future editors in this way be "standard", "above and beyond the call of duty", or "crazy"? How can you profile a Python script? and other coding contests often have a maximum time to run or people boast of how fast their particular solution runs. With Python, sometimes the approaches are somewhat kludgey - i.e., adding timing code to __main__.  What is a good way to profile how long a Python program takes to run? Identify GoogleTv from Android app What the #$@&%*! is that called? Is there a term for an issue or activity that occurs only until the moment that you ask someone else to watch? Is there a term for the coincidence that occurs when an issue ONLY occurs until the moment that you ask someone else to watch or help you resolve it?    For example, having an issue with software until you try to show someone, then it stops being problematic.  Or an issue with the computer/phone that resolves as soon as a help ticket is put in.    The final example is when you have a problem with your car (funny noise, etc) - you hesitate taking it in for weeks because you know it will disappear as soon as the mechanic drives it...then it does.    This is different than a , where the action of applying a debugger, for example, changes the behavior of the activity being watched.  It's also different from a 'works on my machine' scenario, where only the original user has the problem.   What is it called when a problem resolves itself just by the mention of it?  Jinxing the issue? :)  I see this every day supporting my software product -- the users are experienced and are not making mistakes at one time then doing something different when I watch.  The problems also truly resolve (coincidentally) as soon as they ask for help - they do not reoccur.  I guess it's a case of synchronicity?