Is every separable space second countable?  I know that the reverse is true. And this statement is true for metric spaces. But is it true for general topological spaces. I tried to find a counterexample but could not find one.
A toplogical space $X$ is said to be second countable   if there exists a countable basis for the topology.      $X$ is separable if there exist a countable dense subset.   Show that a second countable space always is separable.      Give an example of a space that is separable but not    second countable   The first part was easy and i managed to solve it on my own  Assume that $X$ is second countable and let $B_1, B_2, B_3...$ be a countable basis. Construct the set $A$ by taking one element $x_n$ out of every $B_n$. Then $A$ becomes dense because every open set $U$ contain a $B_n$ which in turn contain the element $x_n$ in $A$.    Again, like in the question i posted yesterday, i had trouble to come up with an example but our professor gave an example during the last lecture but i dont fully understand it:  "For an example of a separable but not second countable space let $X$ be the real line provided with the cofinite topology, i.e. where a set is open if its either empty or if its complement is finite.   Then every infinite set is dense and hence $X$ is separable.  But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  We asked him to explain it in more detail but his english is poor and nobody understood anything...:)       Def: A subset $A$ of a topological space $X$ is dense in $X$ if for any point $x$ in $X$, any neighborhood of $x$ contains at least one point from $A$.   $1.$ "Then every infinite set is dense"  Take $A$ to be one of these infinite sets. If $x \in A \subset X$ then it trivially fulfills the definition. But if we take a $x \not\in A$ why is it not possible to find a neighborhood in these finite sets that does not contain an element in $A$?  edit: its obvious if it is only open neighborhoods, open sets in the topology, but that is not specified in the definition.  $2.$ I am not fully understanding this either  "But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  Is it possible to express this in a simpler way?
The Download CSV link seems broken in SEDE query result window I've created   which works nice. I would really like to download those results as an CSV file. Therefor I click on the provided link on the right hand side of the screen called Download CSV. The result is:  The webpage cannot be found  https://data.stackexchange.com/stackoverflow/csv/202355?UserId=187606 I expected: A file in CSV format. It looks like a regression of  Queries without a parameter work as advertised. Can this be fixed? Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares: $$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$ Let's consider $a,b&gt;0$ for now. $$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$ I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then: $$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$ $$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$ $$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$ $$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$ $$|p-1| \leq 1$$ $$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears: $$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$ Finally, we rewrite: $$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$ Or, using the Pochhammer symbol:  $$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$  My questions are:  Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form? Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it?  If the method I used is correct, it can be used for any integral of the form ($m \geq 2$): $$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$ Two Answers, One Question Sometimes, when I answer a question on Stack Overflow, I have two different answers that are equally valid. This happens most when answering architectural questions where there are many valid ways to solve a problem.  Is it better to put each alternative into a single answer, or put each alternative in its own answer?  Posted each alternative separately will allow the voting system to float the best answer to the top, but I don't want to be accused of rep whoring by "flooding" a question with answers.  What's your opinion? One or two answers? Why can't polynomials have negative exponents or division by a variable Why can't:  $$2x^{-3} - 3x$$  or  $$\frac{1}{2x}$$  be polynomials too?  Why have a definition that excludes these algebraic forms? Wouldn't the thermodynamic cost of creating alternate universes make the Many Worlds interpretation implausible? I was thinking about the many interpretations of quantum physics, and one thing that never made sense to me was the many world's interpretation. Basically at any given moment for which something exists in a superposition, isn't this interpretation basically saying that there would have to exist a new universe for every possible discrete outcome that could result once the measurement is made?   So let's say an electron's location is a superposition of two distinct locations. Doesn't that mean there would at that moment have to be two distinct universes for each distinct location that is superposed? If so, then clearly the amount of universes being spawned every moment must be astronomical.   There could be trillions upon trillions of electrons(for instance) having superposed states, and then every moment in time (I suppose every planck time?), it seems there would have to be yet another set of universes spawning for every superposition.  How could this be? Surely there must be some kind of thermodynamic cost to replicate an entire universe. What would fuel or supply the additional mass and energy for every universe at every moment for every superposed state?  To my lay person's mind, this seems to be a very big weakness of this interpretation, but I needed to ask people with the physics background required to answer this question. Could there be some kind of loophole that could allow alternate universes to spawn so rapidly? Otherwise, doesn't this present a bit of a problem for taking this interpretation seriously? Almost half of disk space unaccounted for on Windows 7 computer I'm trying to determine what's eating almost half of disk on my Windows 7 Lenovo laptop but haven't been able to determine what.  I read some of the answers on this site but none of the causes seems to be the cause of my problem.  What I've done so far:   Checked Windows' System Restore points but only minimum use, either way I deleted around 2GB if usage from there Used WinDirStat and it came back with a total of 71.3GB usage at the root level C Drives properties says 135GB being used out of 138 GB total, ie, 135 GB - 71.3GB = 63.7GB missing somewhere No "Unknown" files reported by WinDirStat Windows Disk Management utility doesn't report any unused partitions or unclaimed space chkdsk comes out clean Windows disk cleanup tool doesn't report anything big, just small stuff in the lower MB range Lenovo backup tool hasn't been used nor there is a folder for the backup images   Any idea on what to look at and/or do?  EDIT: some additional info... My advisor wants me to include a name of someone who has no contribution to the paper My first advisor wants me to include my second advisor's name in a paper that I worked on before she (2nd advisor) came along.   Should I include her name even though she didn't do anything in that paper? I don't think she even read it. Can I cross the USA-Canada border with a birth certificate and a passport locator number? My US passport took too long to be shipped to me, and it will arrive after I need to leave. Will I be able to cross the border from the United States to Canada with a birth certificate (copy) and a passport locator number? How to completely disable the Windows taskbar? I've installed Taskbar eliminator, while it works most of the time, alerts being the taskbar back.  I've also installed object dock which can disable the taskbar, object dock appears during window changes.   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