Let $S$ be a half-open interval $(0,1]$. If we define $d$ on $S$ by  $$d(x,y) := \left|{1\over x} - {1\over y}\right|\;,$$  then show that $d$ is a metric on $S$. Also, prove that $(S,d)$ is a complete metric space.  I think that this problem would begin by showing the $3$ properties of a metric space. And then show that it is Cauchy?
Let $X=(0,1]$ and $d(x,y)=\left|\frac{1}{x}-\frac{1}{y}\right|$. I've proven $(X,d)$ is a metric space but I don't know how to show its completeness. How can I do that?
Random piece of plastic I have recently purchased a hybrid bike with suspension (not sure if that is relevant to this question). It came with two pieces of plastic. Any ideas what they may be for? When I create a form using the form API, the resulting html includes all sorts of unwanted divs. For example, if I use:  $form = array(     'myFieldset' =&gt; array(         '#type' =&gt; 'fieldset'     ) );   The resulting html looks like this:   &lt;form action="/drupal-7.15/?q=chercheur" method="post" id="form-chercheur" accept-charset="UTF-8"&gt;     &lt;div&gt;         &lt;fieldset class="form-wrapper" id="edit-myfieldset"&gt;             &lt;div class="fieldset-wrapper"&gt;             &lt;/div&gt;         &lt;/fieldset&gt;         &lt;input type="hidden" name="form_build_id" value="form-87Dv5ezVpaLI5mD345ofMkLgPhsv_dawudUp73ePQTc" /&gt;         &lt;input type="hidden" name="form_token" value="sXXINWr7rmParm0tRJoEb6QKGIH_QH-ffRSOtnvxYY0" /&gt;         &lt;input type="hidden" name="form_id" value="form_Chercheur" /&gt;     &lt;/div&gt; &lt;/form&gt;   As you can see, two extraneous divs are added here. They are making the javascript I'm writing uselessly complicated, so I was wondering if there was a way to remove the extraneous divs. Ideally, I would like to obtain something like this:  &lt;form action="/drupal-7.15/?q=chercheur" method="post" id="form-chercheur" accept-charset="UTF-8"&gt;     &lt;fieldset class="form-wrapper" id="edit-myfieldset"&gt;     &lt;/fieldset&gt;     &lt;input type="hidden" name="form_build_id" value="form-87Dv5ezVpaLI5mD345ofMkLgPhsv_dawudUp73ePQTc" /&gt;     &lt;input type="hidden" name="form_token" value="sXXINWr7rmParm0tRJoEb6QKGIH_QH-ffRSOtnvxYY0" /&gt;     &lt;input type="hidden" name="form_id" value="form_Chercheur" /&gt; &lt;/form&gt;   Is this possible? Beta regression of proportion data including 1 and 0 How to highlight source code in HTML? I want to highlight C/C++/Java/C# etc source codes in my website.  How can I do this?  Is it a CPU intensive job to highlight the source code? Rubik's Cube Combination Could anyone explain why the number of legal or reachable combinations of a $3\times 3\times 3$ Rubik's Cube is $1/12\mbox{th}$ of the total. I understood the logic behind the total number of combinations: $8! \cdot 2^{12} + 3^{12} \cdot 12!$. What I am not able to understand is why do we divide this quantity by $2 \cdot 3\cdot 2$. I don't have any idea about parity or group/set theory. Is there another possible explanation for it? This question is essentially the same as one of my , except the only difference is that I am using "only" to change the text at the same location (i.e., ). In other words, I am destroying my earlier text as I go, and not adding new ones underneath the previous ones.   Since the text for the first "only" is shorter than the second "only", therefore the image will shift location. I would like for the image to remain at its location.   Here is an example     it may not be obvious but the second image is higher up than the first.  The code is here, can someone find a solution?  \documentclass{beamer}  \usepackage{graphicx}  \title[Title]{Presentation} \author{Sandro Botticelli} \institute{Italy} \date{1484}  \begin{document}   \begin{frame} \begin{figure} \includegraphics[scale = 0.8]{Venus}     \caption{Nascita di Venere} \end{figure}  \only&lt;1&gt;{In the centre the newly-born goddess Venus stands nude in a giant scallop shell. Its size is purely imaginary, and is also found in classical depictions of the subject. }    \only&lt;2&gt;{Alternative identifications for the two secondary female figures involve those also found in the Primavera; the nymph held by Zephyr may be Chloris, a flower nymph he married in some versions of her story, and the figure on land may be Flora. Flora is generally the Roman equivalent of the Greek Chloris; in the Primavera Chloris is transformed into the figure of Flora next to her, following Ovid's Fasti, but it is hard to see that such a transformation is envisaged here.} \end{frame}   \end{document} Can Dropbox be used with multiple accounts on one computer? If I have a PATH variable that (when exploded onto multiple lines) contains something like this (with a few notes added by me):  /Users/brandon/.rvm/gems/ruby-2.0.0-p353/bin /Users/brandon/.rvm/gems/ruby-2.0.0-p353@global/bin /Users/brandon/.rvm/rubies/ruby-2.0.0-p353/bin /opt/local/bin /opt/local/sbin /usr/local/sbin /usr/local/bin   # &lt;&lt;--- notice this path appears again later /usr/bin         # &lt;&lt;--- notice this comes AFTER /usr/local/bin /bin /usr/sbin /sbin /usr/local/bin   # &lt;&lt;--- here it is again! /usr/local/git/bin /Users/brandon/bin /Users/brandon/.homebrew/bin /usr/local/mysql/bin /usr/local/share/npm/bin /Applications/Postgres93.app/Contents/MacOS/bin /Users/brandon/.rvm/bin   is it possible for the presence of the second /usr/local/bin/ to cause the /usr/bin to take precedence over /usr/local/bin?  This seems crazy (to me: I've never heard of such a thing) but that's exactly the behavior I'm seeing.  I've just installed PHP with Homebrew, and yet the system PHP (from /usr/bin) is taking precedence:  $ which php /usr/bin/php   BUT, after manually altering the PATH (rather than digging through all the files sourced by my .zshrc to try to figure out why &amp; where the duplication is coming from) to remove the second instance of /usr/local/bin, I find that /usr/local/bin is taking precedence over /usr/bin as I would have expected it to in the first place:  $ which php /usr/local/bin/php   What would cause this?  Is this normal? Does scikit-learn have forward selection/stepwise regression algorithm? I'm working on the problem with too many features and training my models takes way too long. I implemented forward selection algorithm to choose features.  However, I was wondering does scikit-learn have forward selection/stepwise regression algorithm? How do you remove the axis multiplier? A chart with data in the range of 200,000 and small difference between max and min values, will be assigned a 10^5 multiplier to the entire axis. This results in the ticks becoming meaningless, since all become 2.   Example:-  \begin{tikzpicture} \begin{axis}   \addplot coordinates {   (100,200001)   (200,200003)   (300,200005)   };     \end{axis} \end{tikzpicture}   When this is plotted the y-axis has 5 ticks, all of which are 2. The y-axis has a multiplier of 10^5. This has effectively removed all information from the graph.  I may use the below to set the explicit value of the tick values but this does not implement change in the display "style" of the axis ticks.  ytick = {200000, 200005, 200010}   I have also tried increasing the precision of the tick label to 5. However this only results in the graph showing labels such as 2.00000, 2.00005, 2.00010 with a 10^5 axis multiplier.  yticklabel style={precision = 5,}   My question is "How do I remove/change this 10^5 multiplier, such that the y-axis ticks are displayed in the form 200000, 200005, 200010. How to test whether a set of four points can form a parallelogram Given four points $(x_1,y_1)$,$(x_2,y_2)$,$(x_3,y_3)$,$(x_4,y_4)$. How can we efficiently test them to make sure whether they are vertices of a parallelogram?  I think evaluating and comparing the distance between two points seems to be inefficient. What do you think? Do you have any better method? If the matrix $A^3 = 0$, is $A$ singular? The validation tab in your script tool properties allows you to put parameters in categories. The :     Put parameters in different categories. Categories are expandable   window shades with a collection of parameters.     I am using the categories to organize my parameters, so I don't care about the expandable/collapsible feature;  it just adds a lot of time to the user.  This is all within a normal toolbox, not a script toolbox.  v10.1    Does anyone know how to have these categories expanded by default?  Here is my only validation code.  I am only utilizing categories.  The other parts are the default.  def initializeParameters(self):     """Refine the properties of a tool's parameters.  This method is     called when the tool is opened."""     self.params[0].category = "General"     self.params[1].category = "General"     self.params[2].category = "Inudaciones"     self.params[3].category = "Cienegas"     self.params[4].category = "DrenajesPrincipal"     self.params[5].category = "DrenajesSensillos"     self.params[6].category = "Inudaciones"     self.params[7].category = "Cienegas"     self.params[8].category = "DrenajesPrincipal"     self.params[9].category = "DrenajesSensillos"     return Is this correct?     That is a real cool answer.   I learned that that was incorrect, since "real" is an adjective which can describe a noun, e.g. "real answer" but it is not an adverb which can describe an adjective, "real cool". Instead you would have to say:     That is a really cool answer.   Since "really" is an adverb. Where can I find PSD file or tutorial to create the effect of frame with shadow (see red arrow in image) Why are there holes in the icing? I tried to follow the new Blender 2.8 tutorial from Blender Guru.   When entering "Sculpting" with the icing there were suddenly alot of holes in it. Why is this? iCloud Disassociation for a Mac Computer I had a Mid-2012 MacBook Pro that recently stopped booting. I narrowed this down based on troubleshooting of the boot device to likely the SATA/hard drive connector/cable no longer working for whatever reason.  I have decided to sell the system. I'd like to know if I remove the hard drive (or in this case, an after-market SSD I've installed) from the system, will there be any other data (i.e. iCloud data) stored on the system on any other chip that I can reset in any way? I plan to reset the SMC/NVRAM before I sell the system off of to someone else.  Additionally, if I remove the hard drive, is the system still associated to iCloud (specifically Find My Mac) or any other service? If so, how do I disassociate it? The system will no longer boot even if I put in my boot device.  Thanks! How to prove $\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$ If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\int_{-\infty}^{+\infty} f(x) \, dx$ exists. How can I prove that $$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx\text{ ?}$$ Is the claim that functions of independent random variables are themselves independent, true?   I have seen that result often used implicitly in some proofs, for example in the proof of independence between the sample mean and the sample variance of a normal distribution, but I have not been able to find justification for it. It seems that some authors take it as given but I am not certain that this is always the case. Here is the problem setup  \begin{equation} \begin{array}{c} \text{min} \hspace{4mm}  \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} \\ s.t \hspace{5mm} \mathbf{b} \in \mathbb{R}^{N} \\ \hspace{9mm}b_0=1\\ \hspace{62mm}b_i=b_{i-2}b_{i-1} \hspace{5mm} \forall \hspace{5mm} 2&lt;i\leq N-1 \end{array} \end{equation}  where $\mathbf{A}$ is a $N\times N$ positive semi-definite matrix  To handle $b_0=1$ constrain, I know I can introduce Lagrange multiplier in objective function  \begin{equation} \begin{array}{c} \text{min} \hspace{4mm}  \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} + \lambda(\mathbf{b}^{T}_{}\mathbf{u}^{}_{}-1) \end{array} \end{equation}  where $\mathbf{u}_{}^{}$ is a $N\times 1$ column vector and $u_0=1$ and $u_i=0 \forall 1&lt;i\leq N-1$.  But I dont know how to handle second set of constraints?   EDIT: Example  Let $N=4$. Therefore I am looking for $\mathbf{b}=[b_0 \quad b_1 \quad b_2 \quad b_3]^T = [1 \quad b_1 \quad b_2 \quad b_1b_2]^T$. How do I include $b_3=b_1b_2$ constraint in objective function