So if a certain sequence $a_n$ is convergent then its bounded.So from Bolzano-Weierstrass $a_n$ has a convergent sub-sequence, but where do I continue from here?
Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.  I also looked up these related questions - , ,  - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.      Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.   Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.   There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.   Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$.   For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.    $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$ $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$   The one which does forms a monotone subsequence.
I'm using using algorithm2e package with the following settings:  \usepackage[ruled,vlined,linesnumbered,noresetcount]{algorithm2e}   But for some subset of algorithms in the same document, I want to use this setting:  \usepackage[boxed]{algorithm2e}   Is there a way to override the default setting just for a subset?  I found a similar question:  Here newenvironment is used to override the default setting of an option. But I want to override the option itself.  I wish there was a way to load a package several times with different options for each instance. I would like to know how to understand/analyse the circuit below. And sorry but my level in electronics is not that high and I could only tell that the opposing transistors on the left are meant to cancel any current from the input , yet I don't really understand the B stage which I think the C stage is well related to , and if I wanted to analyse the nodes around the circuit apart from (A,B,C) How can I do it ?  Is it by measuring the current on resistors/nodes?  and note that the output voltage of the circuit in node C is zero this is what it meant to be.I would appreciate any help.  Regards Mongo find documents where array contains x values of given array I'm trying to draw a rounded rectangle with round "holes" using  but I need the rectangle borders to be thick and "holes" borders to be thin. Is there a better way to do so than:  \begin{tikzpicture}[x=1mm, y=1mm]     % Background         \fill[fill=black!10] (1.5, 1.5) rectangle (15, -15);      % Fill     \fill[fill=black!2, even odd rule, rounded corners=0.5mm]          (0, 0) rectangle (10, -10)          (2, -2) circle[radius=1.25mm]         (2, -5) circle[radius=1.25mm];     % Rectangle border     \draw[thick, rounded corners=0.5mm] (0, 0) rectangle (10, -10);     % Holes borders     \draw[thin] (2, -2) circle (1.25mm);     \draw[thin] (2, -5) circle (1.25mm); \end{tikzpicture}   The result image should look like this: Are two norms equivalent if they induce the same topology on a vector space? Omission of definite article with musical instruments What is the rule for omitting/including the definite article in the following sentences:     I used to play piano.   I used to play the piano.   I would pick the first sentence, but I've heard people say the second sentence even when they are not referring to a specific piano. 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