Where $c = 2^{\aleph_0}$.  I have no idea how to tackle this problem. Any help is greatly appreciated!
The book Theory of Functions of a Real Variable by I. P. Natanson, proves that a denumerable or finite union of pairwise disjoint sets of cardinality $c$ has cardinality $c$.  The proofs given in the book are fairly easy, using the axiom of choice. But then it is left as an exercise a generalization of the above theorems for not necessarily pairwise disjoint sets.  I have searched all over the web, but all the proofs found so far are for pairwise disjoint sets.  The proof given in the book relies on the fact that each set $A_k$ can be matched with a set like $[a_{k-1},a_k)$, with $a_k$ point of a partition of $[0,1]$, such that $\bigcup_{k=1}^{\infty\lor n}A_k=[0,1)$. So it is easy to guess the bijective function, but I can't generalize this proof.  Any help is highly appreciated. Thanks and regards.
Why does 'g' sometimes read as 'j'? Most words with 'g' read as 'g'   Geek Goose Goods Goal   But sometimes 'g' reads as 'j'   Gymnastics Genesis Genius Genetic   And in the middle of a sentence, compare:   Log Logistics   Or   Frag Fragile For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge? For what values $s$ does $\int_{0}^{\infty} \frac{\sin x}{x^s}dx$ converge?  Solution says $0&lt;s&lt;2$.  Any ideas? short view distance in 3d view? There is no or there was no I read a sentence in an exam:     There is no child who did not bring his parents to the picnic.    I think that the sentence does not have tense-agreement. Why is is used instead of was? Please explain it to me. Why does Ubuntu not allow the GUI users to login as root at the time of system start up like administrator in Windows?  Why they are restricting us to login only through the terminal? How do I unlock a SQLite database? sqlite&gt; DELETE FROM mails WHERE (`id` = 71); SQL error: database is locked   How do I unlock the database so this will work? Example where convergence in mean does not imply convergence in quadratic mean Is there an example where convergence in mean does not imply convergence in quadratic mean. $E|X_n-X|\rightarrow 0$ but $E|X_n-X|^2 \nrightarrow 0$.  It seems like this should be fairly straightforward, but I cannot really come up with an answer.  I'd imagine the answer would look something like   $$X_n=\begin{cases}a &amp; \text{with probability } \frac{1}{n} \\\  0&amp; \text{with probability } 1-\frac{1}{n} \end{cases}$$ Cloning an SD card onto a larger SD card I'm using Ubuntu 12.04. I have an old 4GB SD card, and have just bought a new 16GB SD card. I would like to copy everything from my old 4GB SD card onto the new 16GB SD card. I was afraid that the usual copy-and-paste trick would miss out something, so I wanted to clone the old 4GB SD card onto the 16GB SD card. I used the dd command in a shell, following instructions in .  This worked with one small caveat. The new SD card now appeared to be 4GB. I later found out this is because the primary partition on the 16GB SD card has now shrunk to 4GB. I have solved this problem, and I think I understand it. However, my question now is, how do I clone my old 4GB SD card onto my new 16GB SD card without making the primary partition on the 16GB SD card shrink to 4GB? Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\in\Bbb Z$ Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers.   I tried to do it by comparing the two expressions but I can't proceed. How to show that the determinant of the following $(n\times n)$ matrix  $$\begin{pmatrix} 5 &amp; 2 &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ 2 &amp; 5 &amp; 2 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; 2 &amp; 5 &amp; 2 &amp; 0 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots&amp; \vdots&amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots \\ 0 &amp; \cdots &amp; \cdots &amp; 0 &amp; 2 &amp; 5 &amp; 2 \\ 0 &amp; \cdots &amp; \cdots &amp; \cdots &amp; \cdots &amp; 2 &amp; 5 \end{pmatrix}$$  is equal to $\frac13(4^{n+1}-1)$?    More generally:     How does one compute the determinant of the following  (where the three diagonals are constant)?      $$M_n(a,b,c) = \begin{pmatrix} a &amp; b &amp; 0 &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ c &amp; a &amp; b &amp; 0 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; c &amp; a &amp; b &amp; 0 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots&amp; \vdots&amp; \vdots &amp; \vdots&amp; \vdots &amp; \vdots \\ 0 &amp; \cdots &amp; \cdots &amp; 0 &amp; c &amp; a &amp; b \\ 0 &amp; \cdots &amp; \cdots &amp; \cdots &amp; \cdots &amp; c &amp; a \end{pmatrix}$$      Here $a,b,c$ can be taken to be real numbers, or complex numbers.   In other words, the matrix $M_n(a,b,c) = (m_{ij})_{1 \le i,j \le n}$ is such that $$m_{ij} = \begin{cases} a &amp; i = j, \\ b &amp; i = j - 1, \\ c &amp; i = j + 1, \\ 0 &amp; \text{otherwise.} \end{cases}$$  There does not seem to be an easy pattern to use induction: the matrix is not a diagonal  of the type $M = \bigl(\begin{smallmatrix} A &amp; C \\ 0 &amp; B \end{smallmatrix}\bigr)$ (where we could use $\det(M) = \det(A) \det(B)$ for the induction step), and there are no lines or columns with only one nonzero entry, so  gets complicated quickly.  Is there a general pattern that one could use? Or is the answer only known on a case-by-case basis? It's possible to compute the determinant by hand for small $n$:  $$\begin{align} \det(M_1(a,b,c)) &amp; = \begin{vmatrix} a \end{vmatrix} = a  \\ \det(M_2(a,b,c)) &amp; = \begin{vmatrix} a &amp; b \\ c &amp; a \end{vmatrix} = a^2 - bc \\ \det(M_3(a,b,c)) &amp; = \begin{vmatrix} a &amp; b &amp; 0 \\ c &amp; a &amp; b \\ 0 &amp; c &amp; a \end{vmatrix} = a^3 - 2abc \end{align}$$  But there is no readily apparent pattern and the computation becomes very difficult when $n$ gets large. Getting arguments underneath \min in an align environment Illogical GTK theme behaviour on Ubuntu 18.04 How to construct a relative path in Java from two absolute paths (or URLs)? declare property as object? Alternative version of the Final Value theorem for Laplace Transform Partitioning $\{1,2,\ldots,k\}$ into $p$ subsets with equal sums, where $p$ is prime Let $p$ be a prime natural number. For which positive integer $k$ can the set $\{1,2,\ldots,k\}$ be partitioned into $p$ subsets with equal sums of elements ?  Obviously, $p\mid k(k+1)$. Hence, $p\mid k$ or $p\mid k+1$. All we have to do now is to show a construction. But I can't find one. I have tried partitioning the set and choose one element from each set but that hasn't yielded anything. Any hint will be appreciated. Can anyone identify this 90s teen-fantasy series, gems in backbone, big magic, evil king? Non-principal ideal in $K[x,y]$? Let $K$ be a field. Then $K[x,y]$ is not a PID because $x$ is irreducible but the quotient $K[x,y]/(x)$ is isomorphic to $K[y]$, which is not a field.  So there must be an ideal in there which is not principal. What is an example of such an ideal? $ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $ $$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $$   (can this be duplicate? I think not) I tried it using many methods  $1.$ Solve this conventionally taking $1^\infty$ form in no luck  $2.$ Did this, expand $ {(1+x)^{\frac1x}}$ using binomial theorem got $\frac13$ then grouped coefficients of $x^0$ and it cancelled with $e$ then took coefficient of $x$ cancelled with $\frac{ex}{2}$ and so on   very messy right ? at last I got $\frac13$ but that's not the expected answer!  I must have went wrong somewhere can anyone help me with this. Let $\mu$ be a general measure, suppose $f,g$ has compact support on $\mathbb{R}$, when does the integration by parts formula hold $$\int f'g d\mu = - \int g'fd\mu?$$ I know in general this is false, we can take $\mu$ to be supported on a point, say $0$, then it is not necessarily true that  $$f'(0)g(0) = -g'(0)f(0).$$  If $\mu$ is absolute continuous w.r.t. Lebesgue measure, we have $\frac{d\mu}{dx} = h$ $$\int f'gd\mu = \int f'gh dx = -\int f(gh)'dx$$ where $(gh)'dx$ might be a measure. but we can not recover the form $\int g'fh dx$.  Thank you very much!