Are $A^c$ and $B^c$ homeomorphic?
Are the complements of two homeomorphic compact, connected subsets of $\mathbb{R}^2$ homeomorphic?
Is there any difference between like best and like the best? Is there a difference between these two sentences? Are they the same?   What color do you like the best? What color do you like best?   Is there a difference in their meaning? The preimage of continuous function on a closed set is closed. Isomorphism between $\mathbb{C}^* $ under multiplication with $\mathbb{C}$ under addition. Does there exists an Isomorphism between $\mathbb{C}^* $ under multiplication with $\mathbb{C}$ under addition.  Attempt In my opinion It does not exists.  If there exists such an isomorphism say $\phi :\mathbb{C}^* \to \mathbb{C}$ then, from the fact that Isomorphism takes identity to identity we have $$\phi \left( 1 \right)=0$$  From the fact      Given $\phi:G \to H$ be an isomorphism, then For a fixed integer $k$ and a fixed group element $b$ in $G$, the   equation $x^k=b$ has the same number of solutions in $G$ as does   the equation $x^k=\phi\left(b\right)$ in $H$.   We have $x^4=1$ and $x^4=0$ must have same number of solutions, but the solutions to these equations are 4 and 1 only. Stack Overflow has successfully borrowed at least one XBox feature (Achievements).  I'd like to see another feature borrowed as well. Let's say I'm playing Halo online and another player is being a bigot.  I have the option to add him to a list of ignored users.  After doing that, all the racist things he says will never be heard through my headphones anymore. Now, on Stack Overflow, I'd like the ability to add a user to a list of ignored users.  This would hide all of that user's comments, and maybe his questions and answers as well (not sure if hiding questions/answers is necessary). This serves multiple purposes:  It would make it easy for users to avoid other users' flamebait. It would give users a more appropriate response to flamebait. The ability to start flame wars would be reduced, since fewer users would see those comments. Users would now have an incentive to be more diplomatic, as there is now a negative consequence to the behavior (the silent treatment). Administrators could review which users are the most often ignored, which would be a strong indication that someone should be  or even locked out of their account. Administrators wouldn't have to spend as much time dealing with complaints about abusive users, since users would have a better way to respond.    makes a good point in a comment to an answer, that I thought should get more visibility:  I had a problem, reported it and got the response &quot;not bad enough to do anything about, sorry&quot;. Which is kind of true, because the only options available to the mods are the nuclear one (penalty box) and the zero-effect one (do nothing). We need a middle ground option where abuse can be handled for you without also being handled for everyone. How long is it safe to drink coffee after brewed? "Started to work" vs "Started working" The one of solution of $x^4 - 2y^2 = -1$ is $x = 1$ and $y = 1$. However, the solution $(1, 1)$  of $x^4 - 2y^2 = 1$ is failed. We know $x = 1$ and $y = 1$ is small integers and we can check by trail method. In case more solutions are existing or not how to check? What are the solutions of $x^4 - 2y^2 = 1$? What is the best way to clean the sensor on a interchangeable lens camera (mirrorless or digital SLR)? bash: ./program: cannot execute binary file: Exec format error I'm trying to run a program, but an error happens like this:  bash: ./program: cannot execute binary file: Exec format error   The result of file program was:  program: ELF-32-bit LSB executable, ARM, EABI4 version 1 (SYSV), dynamically linked(uses share libs), for GNU/LINUX 2.6.16, not stripped   How can I fix this error?  I'm using Ubuntu 14.04.2 (amd64) with VMware. I also tried with Ubuntu i386, but result was same. What is the limit of $n \sin (2 \pi \cdot e \cdot n!)$ as $n$ goes to infinity? I tried and got this   $$e=\sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{k!}$$ $$n!\sum_{k=0}^n\frac{1}{k!}=\frac{n!}{0!}+\frac{n!}{1!}+\cdots+\frac{n!}{n!}=m$$ where $m$ is an integer. $$\lim_{n\to\infty}n\sin(2\pi en!)=\lim_{n\to\infty}n\sin\left(2\pi n!\sum_{k=0}^n\frac{1}{k!}\right)=\lim_{n\to\infty}n\sin(2\pi m)=\lim_{n\to\infty}n\cdot0=0$$  Is it correct? One-hot vs dummy encoding in Scikit-learn Assume that $Y_0,\ldots , Y_n$ are independent random variables with the following identical distribution: $Y_i=1$ with propability $p$ and $Y_i=0$ with propability $1-p$. Also set $S_0=0$ and $S_n=Y_0+\ldots + Y_n$.  It's quite trivial that $(Y_n)_{n\geq 0}$ is a Markov chain, and I managed to show that $(X_n)=(S_n)_{n\geq 0}$ is also a Markov chain, by showing that  $\mathbb{P}(X_{n+1}=i_{n+1}\,|\,X_0=i_0,\ldots,X_n=i_n)=\mathbb{P}(X_{n+1}-X_n=i_{n+1}-X_n\,|\,X_0=i_0,\ldots,X_n=i_n)=\mathbb{P}(Y_{n+1}=i_{n+1}-X_n\,|\,X_0=i_0,\ldots,X_n=i_n)=\mathbb{P}(Y_{n+1}=i_{n+1}-i_n)=\mathbb{P}(Y_{n+1}+i_n=i_{n+1})=\mathbb{P}(X_{n+1}=i_{n+1}\,|\,X_n=i_n)$  However I am now running into the problem that I have to find whether or not $(X_n)=(S_0+\ldots + S_n)_{n\geq 0}$ is a Markov chain. I suspect that it isn't, but i'm having trouble finding a counterexample. I also need to find if $(X_n)=(S_n, S_0+\ldots+S_n)$ is a Markov chain, but I think this actually might be one, since $X_n$ gives "more information" (so to speak) to $X_(n+1)$. Again, having trouble actually proving that.  I appreciate any help, tips etc you can give me. The Bash man page describes use of ${!a} to return the contents of the variable whose name is the contents of a (a level of indirection).  I'd like to know how to return all elements in an array using this, i.e.,  a=(one two three) echo ${a[*]}   returns  one two three   I would like for:  b=a echo ${!b[*]}   to return the same.  Unfortunately, it doesn't, but returns 0 instead.  Update  Given the replies, I now realise that my example was too simple, since of course, something like:  b=("${a[@]}")   Will achieve exactly what I said I needed.  So, here's what I was trying to do:  LIST_lys=(lys1 lys2) LIST_diaspar=(diaspar1 diaspar2)  whichone=$1   # 'lys' or 'diaspar'  _LIST=LIST_$whichone LIST=${!_LIST[*]}   Of course, carefully reading the Bash man page shows that this won't work as expected because the last line simply returns the indices of the "array" $_LIST (not an array at all).  In any case, the following should do the job (as pointed out):  LIST=($(eval echo \${$_LIST[*]}))   or ... (the route that I went, eventually):  LIST_lys="lys1 lys2" ... LIST=(${!_LIST})   Assuming, of course, that elements don't contain whitespace. The Wikipedia page () proves that $$f(x) = \begin{cases}  \exp(-1/x),  &amp; \mbox{if }x&gt;0 \\ 0, &amp; \mbox{if }x\le0 \end{cases}$$ is a non-analytic smooth function. But I don't understand why the fact that "the Taylor series of $f$ at the origin converges everywhere [where does "everywhere" mean here?] to $0$" implies the Taylor series converges to $0$ when $x&gt;0$? Can someone explain this in greater detail? I guess there should be a gap in my understanding of power series. I've seen a couple other questions with this error message but it seemed that the host and ip in question were ones the OP knew. These two are unrelated to my server.  Where and why is postfix generating this error?     Jul 24 17:42:14 email postfix/smtpd[32277]: warning: hostname kimtake.respectedopinion.com does not resolve to address 62.210.28.114   Jul 24 17:42:14 email postfix/smtpd[32277]: connect from unknown[62.210.28.114]   Jul 24 17:42:14 email postfix/smtpd[32277]: disconnect from unknown[62.210.28.114] The speed of light as it approaches a massive body No matter how fast you go, you will aways perceive the speed of light as constant. Taking that as a fact, the special relativity theory was formulated. Now, for what I understand about general relativity, just by standing here on the surface of the planet under the effect of gravity, my frame of reference is a little different from someone in geostationary orbit, and time goes by a little slower for me.  So, considering me and someone in geostationary orbit, does both of us would perceive light at the same speed? If so, does that means that the light is slower down here than up there? Where will changes and fixes to guides and manuals for 5e be found? I'm interested in a grappler build, among others, and don't know where to look to see if the last bullet for the grappler feat has been changed. The grappler example aside, it should be useful to know for future fixes. Set of limit points of a subset of a Hausdorff space is closed. Let $X$ be a Hausdorff space and $A\subset X$.  Define $A'=\{x\in X\mid x\text{ is a limit point of }A\}$.  Prove that $A'$ is closed in $X$.  Relevant information:  (1.) Every neighborhood of a point $x\in A'$ contains a point $y\in A'$ distinct from $x$ (in fact, in a Hausdorff space, every neighborhood of $x$ contains $\infty$-many points of $A$ distinct from $x$)  (2.) In a Hausdorff space, every sequence has a unique limit.  On first glance, it should be an easy proof, but I've made little progress.  I was planning on showing $\overline{A'}=A'$.  Firstly, $A'\subset \overline{A'}$ trivially.  To show $\overline{A'}\subset A'$, proceed by contradiction.  Assume there exists $x\in\overline{A'}$ such that $x\not\in A'$.  This should yield an easy contradiction but I don't see it.  In particular, I'm unsure if the fact that $x\in\overline{A'}$ implies that there actually exists a sequence in $A'$ converging to $x$.  By (2) we know all sequences have unique limits, but do we know that elements in the closure are limits of sequences?  If this is true, it should yield an easy contradiction.  Any help? Concrete Mathematics - The Josephus Problem I've been going through  and have a question on the inductive proof for the . Recurrent relation:  $J(1) = 1$ $J(2n) = 2J(n) - 1$ $J(2n+1) = 2J(n) + 1$  Closed form hypothesis:  $J(2^m + l) = 2l + 1$, for $m \ge 0$ and $0\le l&lt;2^m$  Inductive Proof (for the even case):  $J(1)=1$ $J(2^m+l)= 2J(2^{m-1}+l/2)-1=2(2l/2+1)-1=2l+1$  How do the authors get from $2J(2^{m-1}+l/2)-1$ to $2(2l/2+1)-1$? I'm a 16-year-old male, and I have been interested in tabletop RPGs for a long time, especially D&amp;D. I sorta dove into D&amp;D really fast and hard by buying the Red box, Dungeon Master Kit, Monsters Vault, Players Handbook, Rules book, and even the Shadowfell Campaign not quite knowing what I'd get myself into. I've done tooonnsss of research and even watched live games on Youtube.   For the most part, I'm basically on my own, My friends are somewhat interested in it and want to play, but expect me to Teach/Dungeon Master the game for them, I'd really wish to do this, but need more insight and expertise. I've played the solo Campaign in the Red box and even thought about joining a local group from a popular gaming store in Phoenix, AZ called Imperial Outpost. But I'm not sure if they would take a newbie like myself. Any advice?