# Custom font sizes in LaTeX

When you’re preparing a report in LaTeX (What is LaTeX ?) and want to change the font size of the title and headings to custom font sizes, add the code given below in your preamble. If you are using LyX, go to Documents->Settings->LaTeX preamble and add the code below. (given below for chapter number font size 18, title font 18, center aligned, section font size 16 and subsection font size 14):

\usepackage{sectsty}
\chapternumberfont{\fontsize{18}{22}\selectfont}
\chaptertitlefont{\centering\fontsize{18}{22}\selectfont}
\sectionfont{\fontsize{16}{20}\selectfont}
\subsectionfont{\fontsize{14}{16}\selectfont}

Note that you need to have the package sectsty (see here or here) installed.

For boxed margins, use the fancybox package in LaTeX. Below is an example for box with rounded edges:

\usepackage{fancybox}
\fancyput(3.25in,-4.5in){%
\setlength{\unitlength}{1in}\fancyoval(7,10.0)}

Add the above in the preamble, and you’re done!

If you want to change the font size of text anywhere else in your report, use the following command (given below for font size 18):

# Completeness of the Real numbers

Cut: A cut in ${ \mathfrak{Q} }$ (The set of rational numbers) is a set ${ \alpha \subset \mathfrak{Q} }$ such that

${ \alpha \neq \Phi }$.
${ \alpha^{c} \neq \Phi }$.
If ${ r \in \alpha }$, ${ s \in \mathfrak{Q} }$ and ${ r \leq s }$, then ${ s \in \alpha }$.
${ \alpha }$ does not have a least element.

Real numbers: The set of real numbers ${ \mathfrak{R} }$ is defined as the set of all cuts in ${ \mathfrak{Q} }$.

If ${ \alpha }$, ${ \beta }$ are cuts, then we say that ${ \alpha \leq \beta }$ iff ${ \beta \subset \alpha }$.

Theorem 1 ${ \mathfrak{R} }$ is complete.

Proof: Let ${ B \subset \mathfrak{R} }$ be bounded below.

Now consider ${ \beta_{1}=\bigcup_{\beta \in B} \beta }$.

We first show that ${ \beta_{1} }$ is a cut.

Since ${ \beta \neq \Phi }$ as it is a cut and there exists atleast one ${ \beta \in B }$, ${ \beta_{1} \neq \Phi }$.

${ \forall r \in \beta_{1} }$, ${ \exists \hat{\beta} \in B }$ such that ${ r \in \hat{\beta} }$.

${ \forall s\in \mathfrak{Q} }$ and ${ s>r }$, ${ s \in \hat{\beta} }$ and hence, ${ s \in \beta_{1} }$.

If ${ \beta_{1} }$ had a least element, say x, then since ${ x \in \beta_{1} }$, ${ x \in \beta }$ for some ${ \beta \in B }$. But for every ${ x \in \beta }$, ${ \exists y \in \beta }$ and hence ${ \exists y \in \beta_{1} }$ such that ${ y since ${ \beta }$ does not have a least element. Hence, ${ \beta_{1} }$ does not have a least element. Therefore, ${ \beta_{1} }$ is a cut.

Now, for every ${ \beta \in B }$, ${ \beta \subset \bigcup_{\hat{\beta}\in B} \hat{\beta} }$. Hence, ${ \beta \subset \beta_{1} }$. Hence, ${ \beta_{1} \leq \beta }$ ${ \forall \beta \in B }$. Therefore, ${ \beta_{1} }$ is a lower bound for B.

Consider any other lower bound, say ${ \gamma }$ for B. Now, ${ \gamma \leq \beta }$ and hence ${ \beta \subset \gamma }$ ${ \forall \beta \in B }$.

Hence, ${ \bigcup_{\beta \in B} \beta \: \subset \gamma }$. This implies that ${ \beta_{1} \subset \gamma }$ and ${ \beta_{1} \leq \gamma }$. Hence, ${ \beta_{1} }$ is the greatest lower bound for B.

We have shown that for any arbitrary set ${ B \in \mathfrak{R} }$, there exists a greatest lower bound. Hence, for every ${ B \in \mathfrak{R} }$, there exists a greatest lower bound. Hence, ${ \mathfrak{R} }$ satisfies the greatest lower bound property. Therefore, ${ \mathfrak{R} }$ satisfies the least upper bound property. This implies that ${ \mathfrak{R} }$ is complete.

$\Box$

# Ordered sets

Order: An order on a set is defined as a relation ${ \leq }$ that is reflexive, transitive and antisymmetric i.e

${ a\leq a }$.
${ a\leq b }$ and ${ b \leq c }$ ${ \Rightarrow }$ ${ a \leq c }$.
${ a \leq b}$ and ${ b \leq a }$ ${ \Rightarrow }$ ${ a = b }$.

A function ${ f(x)}$ is said to be order preserving if ${ a \leq b }$ ${ \Rightarrow }$ ${ f(a) \leq f(b) }$. A set ${ M }$ is said to be ordered or simply ordered if for every pair of elements ${ a }$ and ${ b }$ in M, ${ a \leq b}$ or ${ b \leq a }$. If this is not true for every pair, the set is said to be partially ordered. A mapping is said to be an order isomorphism if ${ f(a) \leq f(b) }$ ${ \Rightarrow }$ ${ a \leq b }$.

Two sets that are order isomorphic to each other are said to have the same order type. If a set is finite with ${ n }$ elements, it has order type ${ n }$. The order type of the set of natural numbers is ${ \omega }$. Hence, a set having order type ${ \omega }$ has power ${ \aleph_{0} }$. But a set having power ${ \aleph_{0} }$ need not have order type ${ \omega }$ since the set can be ordered in many ways (in fact, uncountably many ways) and hence can have uncountably many order types.

We say that ${ a if ${ a\leq b }$ and ${ a\neq b }$.

Well ordered set: A non empty set S is said to be well ordered if every non empty subset of S has a least element i.e. ${ Q\subset S }$ and ${ Q\neq \Phi }$ ${ \Rightarrow }$ ${\exists}$ a ${ \in }$ Q such that ${ a ${ \forall }$ ${ x\in Q }$.

eg: The set of natural numbers is well ordered. The intervals [0,1], (2,100) are not well ordered since not every subset has a least element.