Q: Given the function $f(x)=\sum_{n=1}^\infty f_n(x)$, where $f_n(x)=      \left\{      \begin{array}{lr}        0;   \;\;if  \;x&lt; r_n \\        \displaystyle \frac{1}{2^n}; x\geq r_n      \end{array}    \right.   $  Here $r_n$ is an enumeration of $\mathbb{Q}$. Show that $f(x)$ is contionuous at irrationals and discontinuous at rationals.  Here is my work. I proved that each $f_n(x)$ is continuous at irrationals. But now stuck in proving $f(x)$ continuous at irrationals. Can anyone please help me?
Let $q: \mathbb{N} \to \mathbb{Q}$ be a bijection and denote the image of $k \in \mathbb{N}$ by $q_k$.  Let $f: \mathbb{R} \to (0,1)$, $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} $$  Show that:  1) $f$ is strictly increasing  2) $f$ is continuous at $x$ iff $x \in \mathbb{R} \setminus \mathbb{Q}$.  My work:  Let $x, y \in \mathbb{R}$, $x &lt; y$. Then there exists $q_n \in \mathbb{Q}$ s.t. $x &lt; q_n &lt; y$. Now $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} &lt; \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) $$ so $f$ is strictly increasing.  I think my proof of 1) is ok, but how can I prove 2)?
What is the probability that a man likes pink? Proving $R-S$ contains a prime ideal when $S$ is a multiplicative set I'm mainly trying to prove that      If $0\not \in S\subseteq R$ is a multiplicative subset of a   commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.   Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $\exists a,b \in R$ s.t $ab\in I$, but $a,b \not \in I$. Then looked each case. However, I stuck when $a,b \in R-[S\cup I]$. The following is what I've done for this case;  Proof:  In this case, our assumptions are $a,b\in R-[S\cup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J \cap S \not = \emptyset$.   These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J \cap S \not = \emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $\exists (i\in I, r\in R, s_1,s \in S)$ s.t  $$\frac{i+ra}{s_1} =\frac{1}{s}$$ but this implies $\exists s_2 \in S$ s.t  $$ s_2(is + sra -1 s_1) = 0 \quad \Rightarrow \quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$ ,so $$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$ Observe that by our assumption $ab\in I$, so $(s_1 s)b \in I$.  Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.  Edit  Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck. Asking distant cousins for DNA? In my family we have a DNA contradiction – two cousins (at about the 4th cousin level) took a Y-DNA test, but did not match and had two completely different haplogroups.   Since then, I have traced living cousins with the same surname – the only ones left living that I am aware of – and I would love to pay to have them tested to see if we can figure out where the non-paternity event (NPE) or other event happened (i.e. this new cousin could help us determine which line the NPE occurred in depending on which cousin he matches).   Does anyone have a sample letter, or know the best way to go about asking a distant cousin for DNA?   I have their address and am planning to write a letter, but I can't imagine receiving such a letter, so I am not quite sure how to ask without scaring them off. If your visa is on an expired passport but still valid, can you still use it with your new passport, or does the old passport have to have a recent expiration date? If so, how do you transfer your visa to your new passport? I have been wanting to understand how to find the sum of this series.   $$1^p + 2^p + 3^p +{\dots} + n^p$$  I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.  I can prove the formulas for  $$\begin{align}   \sum_{1}^{n} k^2 &amp;= \frac{n(n+1)(2n+1)}{6}\\  \sum_{1}^{n} k^3 &amp;= \frac{n^2(n+1)^2}{4}\\ \sum_{1}^{n} k^4 &amp;= \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\ \end{align}$$  With mathematical induction. But, beyond that even proofs with mathematical induction are difficult.  I'm interested in learning the theory and the proof behind Faulhaber's formula. What is the knowledge required to understand this proof ? How to properly set up a PDO connection Object renders black Like the subject: short sci-fi story about Earth blockaded by other human colonies.  The best humans have colonized some stellar systems. But they begin to treat humans of earth like inferiors. A war began, colonies won the war and blockaded the earth, but despite this the earth won. Who is the author? Limit of a Piecewise Function defined on Rationals and Irrationals Does there exist a polynomial $p(x) \in \mathbb C[x]$ such that $p(x) \notin \mathbb R[x]$ and $p(x)p(-x)=p(x^2)$ ?  I have noticed that if $a_n$ is the leading coefficient of $p(x)$ then $a_n=(-1)^n $, so the leading coefficient cannot be imaginary; also $p(0)^2=p(0)$ so the constant term also is real. Please help. Thanks in advance. Are vacuum fluctuations really happening all the time? In popular physics articles and even some physics classes I've been to, the vacuum of space is described as being constantly full of quantum fluctuations.  Supposedly, all sorts of particle-antiparticle pairs at all scales are constantly appearing and disappearing.  We end up with a mental image of the vacuum as a roiling, choppy sea with all sorts of things going on, rather than a calm, placid background.  However, the vacuum, being the lowest-energy state of a theory, should be an energy eigenstate—which means it is time-invariant (except for a physically-irrelevant phase factor).  So it seems the vacuum really should not be seen as a dynamic entity with all kinds of stuff happening in it, as we're led to believe.   that     A “vacuum fluctuation” is when the ground state of a system is measured in a basis that does not include the ground state; it’s merely a special case of a quantum fluctuation.   So it sounds as if the existence of vacuum fluctuations is contingent on measuring the vacuum—in particular, measuring something that doesn't commute with energy (such as, I guess, the value of a field at a point).  How much truth is there to the idea that vacuum fluctuations are constantly happening everywhere, all the time?  Is that really a useful way to think about it, or just a myth that has been propagated by popularizations of physics? I am using PHP with Apache on Linux, with Sendmail. I use the PHP  function. The email is sent, but the envelope has the Apache_user@localhostname in MAIL FROM (example nobody@conniptin.internal) and some remote mail servers reject this because the domain doesn't exist (obviously). Using mail, can I force it to change the envelope MAIL FROM?  EDIT: If I add a header in the fourth field of the mail() function, that changes the From field in the headers of the body of the message, and DOES NOT change the envelope MAIL FROM.  I can force it by spawning sendmail with sendmail -t -odb -oi -frealname@realhost and piping the email contents to it. Is this a better approach?  Is there a better, simpler, more PHP appropriate way of doing this?  EDIT: The bottom line is I should have RTM. Thanks for the answers folks, the fifth parameter works and all is well. Realtek RTL8723BE Wi-Fi incredibly weak Why is $\sqrt{-x}*\sqrt{-x}=-x?$ How to forward X over SSH to run graphics applications remotely? I have a machine running Ubuntu which I SSH to from my Fedora 14 machine. I want to forward X from the Ubuntu machine back to Fedora so I can run graphical programs remotely. Both machines are on a LAN.  I know that the -X option enables X11 forwarding in SSH, but I feel like I am missing some of the steps.  What are the required steps to forward X from a Ubuntu machine to Fedora over SSH? Sometimes, others just inform us about things without us asking them, other times they do not do that even if we ask them. I want a verb for when others do not give us information — whether because they do not like to or they just don't give it, but we cannot directly ask them to — but we actively extract the information from them sometimes without them being aware of it.   So the word I am looking for shouldn't be only for cases where we coerce or force others to give us information, and it shouldn't be only for cases where others do not like to give us information.  Example:     I needed that information. But I couldn't ask her for it, or most   probably she would not give to me even if I asked her. So, I ....   the information I wanted.   It could be a phrasal verb and so not just a single word, but I couldn't devise a  blank suitable for all possibilities!   Two verbs I've already had but I think are not suitable:  to give someone the third degree   To prize something from/out of somebody Shapefile layer no longer editable (toggle edit option greyed out) in QGIS? Jquery, changing background image with timer Can I install Android on my non-Android device? I would like to know the purpose of nohup these days. I ask because I read this  Nohup is short for “No Hangups.” It’s not a command that you run by itself. Nohup is a supplemental command that tells the Linux system not to stop another command once it has started. That means it’ll keep running until it’s done, even if the user that started it logs out. The syntax for nohup is simple and looks something like this: nohup sh your-script.sh &amp; Notice the “&amp;” at the end of the command. That moves the command to the background, freeing up the terminal that you’re working in. Nohup works with just about any command that you run in the terminal. It can be run with custom scripts as well as standard system commands and command line utilities.  With using linux distributions such as SLES 11.4, or RHEL/CentOS 7.x, I can  remote log in to linux over the network via SSH run the simple program below, or any other, doing ./a.out &amp; log out of linux; type exit in the putty SSH terminal my sample program below, or any other including bash, csh, or tcsh shell scripts, run to completion just fine without the need for nohup typing exit in putty to close an SSH session, I would think that qualifies as a logout?  Is there any purpose or value with nouhup today?  Is it a kernel &gt;= 3.x thing? For however long I can remember (kernel &gt;= 2.6) I've never used nohup and have always just use &amp; with no problems. Can someone give me a practical scenario where nohup would be used? #include &lt;stdlib.h&gt; #include &lt;stdio.h&gt;  /*     sample C code, to print numbers to a file named zz.tmp     the numbers 0 to 29 over 30 seconds */  int main ( int argc, char *argv[] ) {    FILE *fp;    int i;     i = 0;    fp = fopen(&quot;zz.tmp&quot;, &quot;w&quot; );     while ( i &lt; 30 )    {       fprintf( fp, &quot;%d\n&quot;, i++ );       sleep( 1 );    }    fclose( fp );    return 0; }  mysleep.sh #!/bin/bash  sleep 1000  On Centos 7.7, optiplex pc login with keyboard &amp; mouse, my account is bash. I do nohup ./mysleep.sh and then control-z.  When typing jobs I see [1]+ Stopped nohup ./mysleep.sh A ps -ef | grep mysleep shows that process id existing, but if I close the terminal window via X in upper right corner, that process goes away so it seems nohup is not working?