Is the null set a subset of every set?
Is "The empty set is a subset of any set" a convention?
On the canonical isomorphism between $V$ and $V^{**}$ I have the following configuration, but I have not able to find any documentation on how to set a maximum backup files on date rolling style.  I know that you can do this with size rolling style by using the maxSizeRollBackups.  &lt;appender name="AppLogFileAppender" type="log4net.Appender.RollingFileAppender"&gt;     &lt;file value="mylog.log" /&gt;     &lt;appendToFile value="true" /&gt;     &lt;lockingModel type="log4net.Appender.FileAppender+MinimalLock" /&gt;     &lt;rollingStyle value="Date" /&gt;     &lt;datePattern value=".yyMMdd.'log'" /&gt;     &lt;layout type="log4net.Layout.PatternLayout"&gt;       &lt;conversionPattern value="%d %-5p %c - %m%n"  /&gt;     &lt;/layout&gt; &lt;/appender&gt; Matrix is conjugate to its own transpose Mariano mentioned somewhere that everyone should prove once in their life that every matrix is conjugate to its transpose.  I spent quite a bit of time on it now, and still could not prove it. At the risk of devaluing myself, might I ask someone else to show me a proof? Are there any clones/alternatives for running a Stack Exchange style Q&A site? "Always on Top" Windows with Java In Java, is there a way to have a window that is "Always on top" regardless if the user switches focus to another application?  I've searched the web, and all of the solutions lean to some sort of JNI interface with native bindings.  Truly this can't be the only way to do it?.. or is it? Concave implies subadditive Basic mathematical induction regarding inequalities These are just the examples from my textbook, but I don't think it did not explain well.  One of the problem was to prove the inequality   $$n&lt;2^n$$  for all integers $n$.  I understand we assume if $P(n)$ is true, $P(k+1)$ is true.  Thus $k+1&lt; 2^{k+1}$  But in the text book, they added one more line that I don't understand:  $$k+1&lt;2^k+1 \le 2^k+2^k=2\cdot2^k=2^{k+1}$$  I understand  $$k+1&lt;2^{k+1}$$  But how do I know $2^k+1$ is larger than $k+1$, and smaller than $2^{k+1}$?  I have one another example just in case,  proving $2^n&lt;n!$ using induction.  $$2^{k+1}=2\cdot 2^k&lt;2\cdot k!&lt;(k+1)k!=(k+1)!$$  I understand we should start with $2^{k+1}&lt;(k+1)!$, but  Where did $2\cdot k!$ come from? Textbook says it came from inductive hypothesis, but I still don't get it.  Please advise. Flagging a deleted answer results in a message saying I am flagging too quickly Most commented post should be the first post in the blog WordPress default setting is, most recent post is the first post of the blog.  But, I need to "Most commented post should be the first post in the blog". ( Categories page as well as home page) I like to be thorough, but if you feel confident you can skip the first paragraph.  Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds.  If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.  We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).    Here's what I did:   For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$.  Assume $a^2 = 1$.  We want to show that letting $b^2 = 1$ as well will lead to a contradiction.  So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution?  If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction.  If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction.  Similarly, $ab = b$ won't work either.  So it must be that $ab = c$.  So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.  So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$.  We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution.  From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$.  So $abc$ is not a new solution; it's just one.  At this point I'm stuck.  I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$.  What am I missing? Mathematical expression to form a vector from diagonal elements I would like to know is there any way to express mathematically (using matrices multiplication, addition, etc.) a vector which is formed from the diagonal elements of a matrix. For example, I have a matrix called $\mathbf{M}$ and I want to create a vector  $\mathbf{v}$ such that $\mathbf{v}= \text{diagonal elements of } \mathbf{M} $  Any matrix algebra or operation do that ? Thank you. Shapefile projections not matching up in QGIS? To start off with, I'm very new to QGIS. I'm attempting to place a shapefile with locations for 3 substations on a map of Peniche, Portugal. The locations for the substations are given in latitude and longitude coordinates.   When I start off by importing the substation shapefile, I make sure that both the layer CRS and project CRS are set to EPSG:32629, which gives each substation point the correct coordinates, however the scale is at 1:1, which is not correct. As soon as I then add a raster map of the region or a shapefile of Portugal using the same CRS the position of the substations also becomes incorrect.   I'm sure there's a simple solution I'm not getting, but I've been trying for a while and have had no success. Here's a dropbox link to the files How can Planck units be consistent with conflicting dimensions of mass? I suspect I'm missing something obvious, but I'm coming up blank. I've gotten pretty comfortable with so-called natural units over the years: in doing quantum mechanics/QFT, it's common to set $c = \hbar = 1$ and in GR, it's common to set $c = G = 1$. I'm told that quantum gravity researchers are fond of , where (according to that Wiki page) we set $c = \hbar = G = 1$, along with a couple other fundamental constants. I don't know what to make of this, though: it's dimensionally inconsistent in 3+1 dimensions. That same article gives the dimensions of each constant:  $$\begin{align} [c] &amp;= L T^{-1} \\ [G] &amp;= L^3 M^{−1} T^{−2} \\ [\hbar] &amp;= L^2 M T^{−1} \end{align}$$  Setting $c$ and $G$ to 1 implies $L=T=M$. Setting $c$ and $\hbar$ to 1 implies $L=T=M^{-1}$. Hence, making all three unity is internally inconsistent: mass cannot simultaneously have dimensions of length and reciprocal length. So, how do Planck units work? Proving identities (mod $pq$) using Fermat's little theorem? Prove that $\sum_{n=1}^{\infty}\exp\left(-n^{\varepsilon}\right)$ converges. Displaying attachments on visual force page Can I use my Windows 7 license in a dual-boot configuration? Lines appear on the texture in render I am trying to learn Blender. While trying to render a subdivided plane with an image texture applied to it, the subdivision edges are not removed in the rendered result.  Kindly, advise, why these edges appear in the render. Whitespace Problems With Script and Variables Using 'return' instead of 'else' in JavaScript