I don't understand why A∩F and B∩F are relatively open where Brian Scott commented.  Thanks
18771
Let $\mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $\bigcap\mathscr{F}\ne\emptyset$. Prove that $\bigcup\mathscr{F}$ is connected.   If $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $\bigcap\mathscr{F}$. Then either $x\in A$ or $x\in B$. I don't know where to go from here.
93821
PART 1: I was reading the article  Wikipedia. I couldn't understand this line:       "if the two events are causally connected ("event A causes event B"), the causal order is preserved (i.e., "event A precedes event B") in all frames of reference."     Is this an assumption or a consequence of STR? Please explain.       Note: My question consists of 2 parts this is the 2nd part. Below is a genral version of my previous question question:.   PART 2: Let there be three events $A$,$B$ and $C$ s.t: $C$ is the result of Simultaneous occurrence of $A$ and $B$. In other words $C$ occurs iff $A$ and $B$ are simultaneous. Now as we know in STR any two events separated in space are not simultaneous in different frames. So In some frames $C$ will occur and in some $C$ will not occur which will cause paradox. I tried many thought experiments to make such a paradox but i failed. In all the experiments that i thought i could not break the causality even by breaking the simultaneity because everytime the fact: "all signals move slower than light" preserved the causality.    So why causlity remains preserved always? Is it due to the fact that nothing can move faster than light?
48739
I have the following premise:      $$\forall x (Fx \vee \neg Fx) \rightarrow Ga$$   Why I cannot use UI and get: $(Fa \vee \neg Fa) \rightarrow Ga$ And what is the difference between the first premise and the below premise:     $$\forall x ((Fx \vee \neg Fx) \rightarrow Ga)$$   Thanks in advance.
85561
I've seen questions regarding  and , but this doesn't tell me why this field exists in the first place.  Do any internal processes use it, or are there any future plans for using it?
18214
For usability reasons, for example in , the 'viewed' counter should show as  ...instead of... . It's much harder to decipher that there are a million views on the question just by looking at it.  Some people in the comments pointed out internationalization concerns with it. I do think that since there might be a problem with it, that spaces would probably be a better idea.
37048
Ok first, there's that lugubrious use of the semicolon; not necessary.  Second, I didn't open a frame!  You did!  Alls I tried to do is upload a 5MB animated .gif file via Imgur.    Looking further into this, Imgur actually posts an error when you upload an animated .gif larger than 2MB.    Go figure!
118033
Please look at the image. The right column's text jumbles with some comment text when the comment contains a large amount of white-space inside an inline code block. I guess that's not the design may be the comments should be a little bit more restricted from expanding? What do you think?    Now even kinda worse:
112264
I'm looking at , which was closed as off-topic a few days ago. Just a few minutes ago, the OP radically revised it; it doesn't even resemble the original question anymore. In fact, the two versions only share about seven words and some punctuation. Here's a screenshot of the most recent revision:    So, the question for the community is this: Is this proper, or should it be rolled back?  Edit: I went ahead and rolled it back because it has a couple of answers already (responding to the old version). But I would still like to know the community consensus on this.
106398
I am not sure I understand this I just to an explanation to this question:  If $a, b, c \in \Bbb{Z}^+$ and $a\mid bc$, does it follow that $a\mid b$ or $a\mid c$?
73047
Passage:  by Mark Shepherd, page 146. =  Don't read the entire PDF. Sorry, this passage is hard for me to comprehend, so I don't even know what to quote.   :  Which of the following is an unstated assumption made about rural spaces? (a) They are capable of being moved to urban areas  They represent the principal area of production. (c) They are capable of being contained in arcades (d) They are totally different in character to urban spaces (e) They are specifi cally linked with individual spaces of consumption      (a) INCORRECT. It is stated that rural spaces are capable of being moved to urban areas, so this cannot be said to be an unstated assumption. (b) CORRECT. It is said that ‘the spaces of production are paradoxically rural spaces located within urban environments’, and a number of references are made to new forms of rural spaces. Given that such references feature in a discussion about production, the assumption must be made that rural spaces represent the principal area of production, otherwise the logic of the argument is lost. (c) INCORRECT. It is suggested that it is consumption, not production, which takes place within the arcades. (d) INCORRECT. Although it may be the case, it is not a necessary assumption to make that rural spaces are totally different in character to urban spaces. (e) INCORRECT. No fi xed link between rural spaces and individual spaces of consumption is suggested or assumed.      Why's (a) wrong? Where does the passage state that? Why's (c) wrong? Does the passage equate 'rural spaces' with production? Why's (d) wrong? Why's that assumption unnecessary?   This article is too abstract, so please enlarge on and simplify why (b) is right?  What's the logic of the argument?
59959
Why the disparity? And why use 'No.'? Is it from the French?  And the hash or pound sign seems a weird choice too, is there a history or any reason involved?
49252
I'm studying topology since a few months ago and I have never caught a good intuition of the topological spaces, but now I think that I did.  My intuition is the next; as many people point out the open sets capture a notion of nearness, (but I did never understand the exact way of this nearness intuition),the points in the open sets are near, and two points in distinct open sets are far, and then the axioms try to capture the behaviours of the opens.  To find a good intuition I have worked with two examples of spaces which I named A and B; A=[{a,b,c,d,e,f},({a,b,c,d,e,f},{},{a,b,c,e,f},{e,f},{a,b},{a,b,d,e,f},{a,b,e,f})], B=[{a,b,c,d,e,f},({a,b,c,d,e,f},{},{a,b,c,e,f},{a,b,d,e,f},{a,b,e,f})] and I obtained the next pictures (and here is my first question: are these  the correct spatial representaition of the spaces? A is in the right, B is the left, the black circles represent points):                                                                                                 And then the union axiom allows to establish different degrees of nearness, specifically the opens that aren't union of other opens  have near points, and the unions have points nearer than other points that aren't. Concretely in the space A; {e,f} and {a,b} are nearer than c and d, then the unions allow to speak of a "global " nearness.  The intersection axiom also talk about  different degrees of nearness, in the way that if we have near points and we have other near points that have common points, then than points are near, and in fact they are nearer, and so the intersection  talk of a "local" nearness.  And finally the axiom of the necessary membership of the space itself and the empty set, guarantees the fact that "there are nothing outside the space" or the totality of the space and in that sense all the points are near in the space.  Is this a good intuition? if not, please give me a good one.
71765
I would be very glad if you helped me solving this problem. It's about topology. Okay let's begin :    Let $A$ be a non-empty open subset of $\mathbb R$. Let $x_0 \in A$ and $B = [x_0,\infty) ∩ A^c$ (complement of $A$)  First I should prove that $B$ is a non-empty, closed set. Then I should prove that :     $B$ is bounded below.   $\inf B &gt; x_0$   $\inf B$ doesn't belong to $A$.   $[x_0, \inf B) \subseteq A$.    Then I should explain how to deﬁne a number $\beta$ so that $(\beta,\inf B) \subseteq  A$, where $\beta$ doesn't belong to $A$.    I really can't get up with proofs for these question. Can you please help me solving this or at least give me some hint or sketch? I would be very grateful. Thank you :)
36917
Why did MS originally make the decision to maintain these two separate core libs? Maybe they had some scalability issue in mind, but nowadays I never see an application, of any type, that doesn't need both. Does anyone have any inside information on this? It's not really important, but been on my mind for years.  PS. I know what's in the two libs, I know the difference - I'm a big fan of  :) Just wondering what practical use the separation of the two has.
23938
So this is a piecewise defined function I have that I need to talk about:     f(a,b) = \left\{      \begin{array}{lr}        \text{open} &amp; : \text{RMSD}_\text{s-open}\ge6, \text{RMSD}_\text{closed}\ge6\\        \text{closed} &amp; : \text{RMSD}_\text{closed}\le2 \\        \text{semiopen} &amp; : \text{RMSD}_\text{s-open}\le2\\        \text{transition} &amp; :f(a,b)\notin\{\text{open}, \text{closed}, \text{semiopen}\} \\       \end{array}    \right.   So what's the way to align up the colons (conditions) so that it looks all nice and pretty?   Also, if I want to describe transition to be the value of the function when none of the conditions of the previous 3 are satisfied, is the way I wrote it out with f(a,b) not belonging to the set {open,closed,semiopen} a good way of writing it? seems really unprofessional...
7037
My book says:     For most of the small objects, both are same. But for mammoth ones, they are really different ones. And in a gravity-less environment, COG is absent; COM still exists.   Ok, what's the big deal when things are small and big? How these two :  &amp; ?
100108
I have a statement,     Either p or q   and I have to write it in terms of logical connectives but I don't get which logical connector should I be using? Here is what I did (I think there could have been a better way to do this)     $(p \lor q ) \land (\neg((p \Rightarrow q) \land (q \Rightarrow p)))$