1 files changed, 1427 insertions, 32 deletions

diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx
index 28f0397..2b6057d 100644
--- a/šola/ana1/teor.lyx
+++ b/šola/ana1/teor.lyx
@@ -2849,7 +2849,20 @@ Lahko se zgodi,
 \end_layout
 
 \begin_layout Theorem*
-Naj bo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kmoz}{Konvergenca monotonega in omejenega zaporedja}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
 \begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
 \end_inset
 
@@ -3862,7 +3875,20 @@ s čimer dobimo zgornjo mejo
 \end_inset
 
 ,
- je torej zaporedje omejeno in ker je tudi monotono po prejšnjem izreku konvergira.
+ je torej zaporedje omejeno in ker je tudi monotono po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{prejšnjem izreku}
+\end_layout
+
+\end_inset
+
+ konvergira.
 \end_layout
 
 \end_deeper
@@ -7157,7 +7183,7 @@ TODO XXX FIXME PREVERI ŠE V profesrojevih PDFJIH,
 Ideja:
  Izdelati želimo formulacijo,
  s katero preverimo,
- le lahko z dovolj majhno spremembo 
+ če lahko z dovolj majhno spremembo 
 \begin_inset Formula $x$
 \end_inset
 
@@ -7455,7 +7481,20 @@ okolice
 
 \end_deeper
 \begin_layout Theorem*
-Naj bo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hypertarget{kzsppom}{Karakterizacija zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Naj bo 
 \begin_inset Formula $f:D\to\mathbb{R}$
 \end_inset
 
@@ -8111,6 +8150,329 @@ Naj bo
 \end_layout
 
 \begin_layout Example*
+Kvadratna funkcija 
+\begin_inset Formula $f\left(x\right)=x^{2}$
+\end_inset
+
+ je zvezna.
+ Vzemimo poljuben 
+\begin_inset Formula $a\in\mathbb{R},\varepsilon>0$
+\end_inset
+
+.
+ Obstajati mora taka 
+\begin_inset Formula $\delta>0\ni:\forall x\in\mathbb{R}:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(a\right)\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Podan imamo torej 
+\begin_inset Formula $a$
+\end_inset
+
+ in 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ želimo najti 
+\begin_inset Formula $\delta$
+\end_inset
+
+.
+ Želimo priti do neenakosti,
+ ki ima na manjši strani 
+\begin_inset Formula $\left|f\left(x\right)-f\left(a\right)\right|=\left|x^{2}-a^{2}\right|$
+\end_inset
+
+ in na večji strani nek izraz z 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+,
+ da ta 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ nadomestimo z 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in nato večjo stran enačimo z 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da izrazimo 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ v odvisnosti od 
+\begin_inset Formula $\delta$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Računajmo:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|=\left|x-a\right|\left|x+a\right|$
+\end_inset
+
+.
+ Predelajmo izraz 
+\begin_inset Formula $\left|x+a\right|=\left|\left(x-a\right)+2a\right|\leq\left|x-a\right|+\left|2a\right|$
+\end_inset
+
+,
+ torej skupaj 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\left|x-a\right|\left(\left|x-a\right|+\left|2a\right|\right)$
+\end_inset
+
+.
+ Sedaj nadomestimo 
+\begin_inset Formula $\left|x-a\right|$
+\end_inset
+
+ z 
+\begin_inset Formula $\delta$
+\end_inset
+
+:
+ 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|\leq\delta\left(\delta+\left|2a\right|\right)$
+\end_inset
+
+.
+ Iščemo tak 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ da velja 
+\begin_inset Formula $\left|x^{2}-a^{2}\right|<\varepsilon$
+\end_inset
+
+,
+ zato enačimo 
+\begin_inset Formula $\delta\left(\delta+\left|2a\right|\right)=\varepsilon$
+\end_inset
+
+ in dobimo kvadratno enačbo 
+\begin_inset Formula $\delta^{2}+\left|2a\right|\delta-\varepsilon=0$
+\end_inset
+
+,
+ ki jo rešimo z obrazcem za ničle:
+\begin_inset Formula 
+\[
+\delta_{1,2}=\frac{-2\left|a\right|\pm\sqrt{4\left|a\right|^{2}-4\varepsilon}}{2}=-\left|a\right|\pm\sqrt{\left|a\right|^{2}-\varepsilon}
+\]
+
+\end_inset
+
+Toda ker iščemo le pozitivne 
+\begin_inset Formula $\delta$
+\end_inset
+
+,
+ je edina rešitev
+\begin_inset Formula 
+\[
+\delta=-\left|a\right|+\sqrt{\left|a\right|^{2}-\varepsilon}=\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|=\frac{\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|}{1}=\frac{\left(\sqrt{\left|a\right|^{2}-\varepsilon}-\left|a\right|\right)\left(\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|\right)}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}=\frac{\varepsilon}{\sqrt{\left|a\right|^{2}-\varepsilon}+\left|a\right|}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R},a\in\mathbb{R}\ni:\forall\varepsilon>0:D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Število 
+\begin_inset Formula $L_{+}\in\mathbb{R}$
+\end_inset
+
+ je desna limita funkcije 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+,
+ če 
+\begin_inset Formula $\forall\left(a_{n}\right)_{n\in\mathbb{N}}\subset D\cap\left(a,\infty\right):a_{n}\to a\Rightarrow f\left(a_{n}\right)\to L_{+}$
+\end_inset
+
+ ZDB če za vsako k 
+\begin_inset Formula $a$
+\end_inset
+
+ konvergentno zaporedje s členi desno od 
+\begin_inset Formula $a$
+\end_inset
+
+ velja,
+ da funkcijske vrednosti členov konvergirajo k 
+\begin_inset Formula $L_{+}$
+\end_inset
+
+.
+ Oznaka 
+\begin_inset Formula $L_{+}=\lim_{x\to a^{+}}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)=f\left(a+0\right)$
+\end_inset
+
+.
+ Podobno definiramo tudi levo limito 
+\begin_inset Formula $L_{-}=\lim_{x\to a^{-}}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=f\left(a-0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $D\subset\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $a\in\mathbb{R}$
+\end_inset
+
+ da velja 
+\begin_inset Formula $\forall\varepsilon>0:D\cap\left(a,a-\varepsilon\right)\not=\emptyset\wedge D\cap\left(a,a+\varepsilon\right)\not=\emptyset$
+\end_inset
+
+.
+ Naj bo 
+\begin_inset Formula $f:D\to\mathbb{R}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+ V tem primeru velja 
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo 
+\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
+\end_inset
+
+.
+ Če 
+\begin_inset Formula $\exists f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists f\left(a-0\right)$
+\end_inset
+
+,
+ vendar 
+\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
+\end_inset
+
+,
+ pravimo,
+ da ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v točki 
+\begin_inset Formula $a$
+\end_inset
+
+ 
+\begin_inset Quotes gld
+\end_inset
+
+skok
+\begin_inset Quotes grd
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
+\end_inset
+
+ ne obstaja.
+ Zakaj?
+ Izračunajmo levo in desno limito:
+\begin_inset Formula 
+\[
+\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
+\]
+
+\end_inset
+
+Toda 
+\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Funkcija 
+\begin_inset Formula $f$
+\end_inset
+
+ je na intervalu 
+\begin_inset Formula $D$
+\end_inset
+
+ odsekoma zvezna,
+ če je zvezna povsod na 
+\begin_inset Formula $D$
+\end_inset
+
+,
+ razen morda v končno mnogo točkah,
+ v katerih ima skok.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
 Naj bo 
 \begin_inset Formula $f:\mathbb{R}\setminus\left\{ 0\right\} \to\mathbb{R}$
 \end_inset
@@ -8234,7 +8596,7 @@ Da naš sklep res potrdimo,
 \end_inset
 
  in 
-\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)=1$
+\begin_inset Formula $\lim_{x\to a}f\left(x\right)=\lim_{x\to a}h\left(x\right)\eqqcolon L$
 \end_inset
 
 ,
@@ -8243,7 +8605,32 @@ Da naš sklep res potrdimo,
 \end_inset
 
  in 
-\begin_inset Formula $\lim_{x\to a}g\left(x\right)=1$
+\begin_inset Formula $\lim_{x\to a}g\left(x\right)=L$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $A=A\left(x\right)\coloneqq\max\left\{ \left|f\left(x\right)-L\right|,\left|h\left(x\right)-L\right|\right\} $
+\end_inset
+
+.
+ Velja
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+\begin_inset Formula $g\left(x\right)-L\leq h\left(x\right)-L\leq\left|h\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $L-g\left(x\right)\leq L-f\left(x\right)\leq\left|f\left(x\right)-L\right|\leq A\left(x\right)$
 \end_inset
 
 .
@@ -8251,72 +8638,1080 @@ Da naš sklep res potrdimo,
 
 \end_deeper
 \begin_layout Proof
-TODO XXX FIXME DOKAZ V SKRIPTi
+Posledično 
+\begin_inset Formula $\left|g\left(x\right)-L\right|\leq A\left(x\right)$
+\end_inset
+
+.
+ Naj bo sedaj 
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Tedaj velja 
+\begin_inset Formula $\exists\delta_{1}>0\ni:\left|x-a\right|<\delta_{1}\Rightarrow\left|f\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $\exists\delta_{2}>0\ni:\left|x-a\right|<\delta_{2}\Rightarrow\left|h\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+ Za 
+\begin_inset Formula $\delta\coloneqq\min\left\{ \delta_{1},\delta_{2}\right\} $
+\end_inset
+
+ torej velja 
+\begin_inset Formula $\left|x-a\right|<\delta\Rightarrow\left|g\left(x\right)-L\right|<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Zvezne funkcije na kompaktnih množicah
+\end_layout
+
+\begin_layout Definition*
+Množica 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ je kompaktna 
+\begin_inset Formula $\Leftrightarrow$
+\end_inset
+
+ je zaprta in omejena ZDB je unija zaprtih intervalov.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $K\subset\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj je 
+\begin_inset Formula $f$
+\end_inset
+
+ omejena in doseže minimum in maksimum.
 \end_layout
 
 \begin_layout Example*
-\begin_inset Formula $\lim_{x\to0}\frac{1}{1+e^{1/x}}$
+Primeri funkcij.
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $f_{1}\left(x\right)=\frac{1}{x}$
 \end_inset
 
- ne obstaja.
- Zakaj?
- Izračunajmo levo in desno limito:
+ na 
+\begin_inset Formula $I_{1}=(0,1]$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f_{1}$
+\end_inset
+
+ je zvezna in 
+\begin_inset Formula $\lim_{x\to0}f_{1}\left(x\right)=\infty$
+\end_inset
+
+,
+ torej ni omejena,
+ a 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ ni zaprt.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{2}\left(x\right)=\begin{cases}
+0 & ;x=0\\
+\frac{1}{x} & ;x\in(0,1]
+\end{cases}$
+\end_inset
+
+ ni omejena in je definirana na kompaktni množici,
+ a ni zvezna.
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{3}\left(x\right)=x$
+\end_inset
+
+ na 
+\begin_inset Formula $x\in\left(0,1\right)$
+\end_inset
+
+.
+ Je omejena,
+ ne doseže maksimuma,
+ a 
+\begin_inset Formula $D_{f_{3}}$
+\end_inset
+
+ ni kompaktna (ni zaprta).
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $f_{4}\left(x\right)=\begin{cases}
+x & ;x\in\left(0,1\right)\\
+\frac{1}{2} & ;x\in\left\{ 0,1\right\} 
+\end{cases}$
+\end_inset
+
+.
+ Velja 
+\begin_inset Formula $\sup f_{4}=1$
+\end_inset
+
+,
+ ampak ga ne doseže,
+ a ni zvezna
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Naj bo 
+\begin_inset Formula $K\subseteq\mathbb{R}$
+\end_inset
+
+ kompaktna in 
+\begin_inset Formula $f:K\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+\end_layout
+
+\begin_deeper
+\begin_layout Itemize
+Omejenost navzgor:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzgor omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\geq n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(x_{n_{k}}\right)=\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+\end_layout
+
+\begin_layout Itemize
+Omejenost navzdol:
+ PDDRAA 
+\begin_inset Formula $f$
+\end_inset
+
+ ni navzdol omejena.
+ Tedaj 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists x_{n}\in K\ni:f\left(x_{n}\right)\leq-n$
+\end_inset
+
+ (*).
+ Ker je 
+\begin_inset Formula $\left(x_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeno zaporedje (vsi členi so na kompaktni 
+\begin_inset Formula $K$
+\end_inset
+
+),
+ ima stekališče,
+ recimo mu 
+\begin_inset Formula $s\in\mathbb{R}$
+\end_inset
+
+.
+ Vemo,
+ da tedaj obstaja podzaporedje 
+\begin_inset Formula $\left(x_{n_{k}}\right)_{k\in\mathbb{N}}\ni:s=\lim_{k\to\infty}x_{n_{k}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ tudi zaprta,
+ sledi 
+\begin_inset Formula $s\in K$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $K$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(s\right)=\lim_{k\to\infty}f\left(x_{n_{k}}\right)$
+\end_inset
+
+.
+ Toda po (*) sledi 
+\begin_inset Formula $\lim_{k\to\infty}f\left(s_{n_{k}}\right)=-\infty$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $f\left(s\right)=-\infty$
+\end_inset
+
+,
+ kar ni mogoče,
+ saj je 
+\begin_inset Formula $f\left(s\right)\in\mathbb{R}$
+\end_inset
+
+.
+ 
+\begin_inset Formula $\rightarrow\!\leftarrow$
+\end_inset
+
+.
+ Torej je 
+\begin_inset Formula $f$
+\end_inset
+
+ navzgor omejena.
+ 
+\end_layout
+
+\begin_layout Itemize
+Doseže maksimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\sup_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ravnokar smo dokazali,
+ da 
+\begin_inset Formula $M<\infty$
+\end_inset
+
+.
+ Po definiciji supremuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)>M-\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)>M-\frac{1}{n_{j}}\geq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\max_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Itemize
+Doseže minimum:
+ Označimo 
+\begin_inset Formula $M\coloneqq\inf_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+ Ko smo dokazali omejenost,
+ smo dokazali,
+ da 
+\begin_inset Formula $M>-\infty$
+\end_inset
+
+.
+ Po definiciji infimuma 
+\begin_inset Formula $\forall n\in\mathbb{N}\exists t_{n}\in K\ni:f\left(t_{n}\right)<M+\frac{1}{n}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $K$
+\end_inset
+
+ omejena,
+ ima 
+\begin_inset Formula $\left(t_{n}\right)_{n}$
+\end_inset
+
+ stekališče in ker je zaprta,
+ velja 
+\begin_inset Formula $t\in K$
+\end_inset
+
+,
+ zato 
+\begin_inset Formula $\exists$
+\end_inset
+
+ podzaporedje 
+\begin_inset Formula $\left(t_{n_{j}}\right)_{j\in\mathbb{N}}\ni:t=\lim_{j\to\infty}t_{n_{j}}$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ velja 
+\begin_inset Formula $f\left(t\right)=\lim_{j\to\infty}f\left(t_{n_{j}}\right)$
+\end_inset
+
+.
+ Toda ker 
+\begin_inset Formula $f\left(t_{n_{j}}\right)<M-\frac{1}{n_{j}}\leq M-\frac{1}{j}$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(t\right)\leq M$
+\end_inset
+
+.
+ Hkrati po definiciji 
+\begin_inset Formula $M$
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(t\right)\geq M$
+\end_inset
+
+.
+ Sledi 
+\begin_inset Formula $M=f\left(t\right)$
+\end_inset
+
+ in zato 
+\begin_inset Formula $M=\min_{x\in K}f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $f\left(a\right)f\left(b\right)<0$
+\end_inset
+
+.
+ Tedaj 
+\begin_inset Formula $\exists\xi\in\left(a,b\right)\ni:f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Interval 
+\begin_inset Formula $I_{0}=\left[a,b\right]$
+\end_inset
+
+ razpolovimo.
+ To pomeni,
+ da pogledamo levo in desno polovico intervala 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\left[a,\frac{a+b}{2}\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\left[\frac{a+b}{2},b\right]$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $f\left(\frac{a+b}{2}\right)=0$
+\end_inset
+
+,
+ smo našli iskano točko 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ sicer z 
+\begin_inset Formula $I_{1}$
+\end_inset
+
+ označimo katerokoli izmed polovic,
+ ki ima 
+\begin_inset Formula $f$
+\end_inset
+
+ v krajiščih različno predznačene funkcijske vrednosti.
+ Torej 
+\begin_inset Formula $I_{1}=\begin{cases}
+\left[a,\frac{a+b}{2}\right] & ;f\left(a\right)f\left(\frac{a+b}{2}\right)<0\\
+\left[\frac{a+b}{2},b\right] & ;f\left(\frac{a+b}{2}\right)f\left(b\right)<0
+\end{cases}$
+\end_inset
+
+.
+ S postopkom nadaljujemo.
+ Če v končno mnogo korakih najdemo 
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ da je 
+\begin_inset Formula $f\left(\xi\right)=0$
+\end_inset
+
+,
+ fino,
+ sicer pa dobimo zaporedje intervalov 
+\begin_inset Formula $I_{n}=\left[a_{n},b_{n}\right]\subset\left[a,b\right]=I_{0}\ni:$
+\end_inset
+
+
+\end_layout
+
+\begin_deeper
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset Formula $\forall n\in\mathbb{N}:I_{n+1}\subset I_{n}$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $a_{n+1}\geq a_{n}\wedge b_{n+1}\leq b_{n}$
+\end_inset
+
+,
+ in
+\end_layout
+
+\begin_layout Enumerate
+\begin_inset CommandInset label
+LatexCommand label
+name "enu:različni-predznaki-istoležnih-clenov"
+
+\end_inset
+
+
+\begin_inset Formula $\forall n\in\mathbb{N}:f\left(a_{n}\right)f\left(b_{n}\right)<0$
+\end_inset
+
+.
+\end_layout
+
+\end_deeper
+\begin_layout Proof
+Ker sta zaporedji 
+\begin_inset Formula $\left(a_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ in 
+\begin_inset Formula $\left(b_{n}\right)_{n\in\mathbb{N}}$
+\end_inset
+
+ omejeni in monotoni,
+ imata po 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kmoz}{izreku o konvergenci monotonih in omejenih zaporedij}
+\end_layout
+
+\end_inset
+
+ limiti 
+\begin_inset Formula $\alpha\coloneqq\lim_{n\to\infty}a_{n}=\sup_{n\in\mathbb{N}}a_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\beta\coloneqq\lim_{n\to\infty}=\sup_{n\in\mathbb{N}}b_{n}$
+\end_inset
+
+ in 
+\begin_inset Formula $\alpha,\beta\in I_{0}$
+\end_inset
+
+,
+ ker je 
+\begin_inset Formula $I_{0}$
+\end_inset
+
+ zaprt.
+\end_layout
+
+\begin_layout Proof
+Sledi 
+\begin_inset Formula $\forall n\in\mathbb{N}:\left|\alpha-\beta\right|=\beta-\alpha\leq b_{n}-a_{n}=\left|I_{n}\right|=2^{-n}\left|I_{0}\right|$
+\end_inset
+
+,
+ torej 
+\begin_inset Formula $\lim_{n\to\infty}\left|\alpha-\beta\right|=0\Rightarrow\alpha-\beta=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Ker je 
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna in 
+\begin_inset Formula $a_{n},b_{n},\xi\in I_{0}$
+\end_inset
+
+,
+ sledi 
+\begin_inset Formula $\lim_{n\to\infty}f\left(a_{n}\right)=f\left(\alpha\right)=f\left(\xi\right)=f\left(\beta\right)=\lim_{n\to\infty}f\left(b_{n}\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Po točki 
+\begin_inset CommandInset ref
+LatexCommand ref
+reference "enu:različni-predznaki-istoležnih-clenov"
+plural "false"
+caps "false"
+noprefix "false"
+nolink "false"
+
+\end_inset
+
+ velja 
+\begin_inset Formula $f\left(\alpha\right)f\left(\beta\right)\leq0$
+\end_inset
+
+.
+ Ker pa 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)$
+\end_inset
+
+,
+ velja 
+\begin_inset Formula $f\left(\alpha\right)=f\left(\beta\right)=f\left(\xi\right)=0$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Corollary*
+Naj bo 
+\begin_inset Formula $I=\left[a,b\right]$
+\end_inset
+
+ omejen zaprt interval 
+\begin_inset Formula $\in\mathbb{R}$
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ zvezna.
+ Tedaj 
+\begin_inset Formula $\exists x_{-},x_{+}\in I\ni:\forall x\in I:f\left(x\right)\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ in 
+\begin_inset Formula $\forall y\in\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]\exists x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ ZDB 
+\begin_inset Formula $f\left(I\right)=\left[f\left(x_{-}\right),f\left(x_{+}\right)\right]$
+\end_inset
+
+ ZDB zvezna funkcija na zaprtem intervalu 
+\begin_inset Formula $\left[a,b\right]$
+\end_inset
+
+ doseže vse funkcijske vrednosti na intervalu 
+\begin_inset Formula $\left[f\left(a\right),f\left(b\right)\right]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Dokaz posledice.
+ Naj bo 
+\begin_inset Formula $y$
+\end_inset
+
+ poljuben.
+ Če je 
+\begin_inset Formula $y=f\left(x_{-}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{-}$
+\end_inset
+
+.
+ Če je 
+\begin_inset Formula $y=f\left(x_{+}\right)$
+\end_inset
+
+,
+ smo našli 
+\begin_inset Formula $x=x_{+}$
+\end_inset
+
+.
+ Sicer pa je 
+\begin_inset Formula $f\left(x_{-}\right)<y<f\left(x_{+}\right)$
+\end_inset
+
+.
+ Oglejmo si funkcijo 
+\begin_inset Formula $g\left(x\right)\coloneqq f\left(x\right)-y$
+\end_inset
+
+.
+ Ker je 
+\begin_inset Formula $g\left(x_{-}\right)=f\left(x_{-}\right)-y<0$
+\end_inset
+
+ in 
+\begin_inset Formula $g\left(x_{+}\right)=f\left(x_{+}\right)-y>0$
+\end_inset
+
+ in 
+\begin_inset Formula $g$
+\end_inset
+
+ zvezna na 
+\begin_inset Formula $\left[x_{-}-y,x_{+}-y\right]$
+\end_inset
+
+,
+ torej po prejšnjem izreku 
+\begin_inset Formula $\exists x\in\left[x_{-}-y,x_{+}-y\right]\ni:g\left(x\right)=0$
+\end_inset
+
+,
+ kar pomeni ravno 
+\begin_inset Formula $f\left(x\right)=y$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo 
+\begin_inset Formula $I$
+\end_inset
+
+ poljuben interval med 
+\begin_inset Formula $a,b\in\mathbb{R}\cup\left\{ -\infty,\infty\right\} $
+\end_inset
+
+ in 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ zvezna in strogo monotona.
+ Tedaj je 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ interval med 
+\begin_inset Formula $f\left(a+0\right)$
+\end_inset
+
+ in 
+\begin_inset Formula $f\left(a-0\right)$
+\end_inset
+
+.
+ Inverzna funkcija 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ je definirana na 
+\begin_inset Formula $f\left(I\right)$
+\end_inset
+
+ in zvezna.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\coloneqq\arctan$
+\end_inset
+
+,
+ 
+\begin_inset Formula $I\coloneqq\left(-\infty,\infty\right)$
+\end_inset
+
+,
+ zvezna.
+ Naj bo 
+\begin_inset Formula $y\in f\left(I\right)$
+\end_inset
+
+ poljuben.
+ Tedaj 
+\begin_inset Formula $\exists!x\in I\ni:y=f\left(x\right)$
+\end_inset
+
+ in definiramo 
+\begin_inset Formula $x\coloneqq f^{-1}\left(x\right)$
+\end_inset
+
+.
+ 
+\begin_inset Formula $f^{-1}$
+\end_inset
+
+ obstaja in je spet zvezna.
+\end_layout
+
+\begin_layout Proof
+Ne bomo dokazali.
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Označimo 
+\begin_inset Formula $g=f^{-1}:f\left(I\right)\to\mathbb{R}$
+\end_inset
+
+.
+ Uporabimo 
+\begin_inset ERT
+status open
+
+\begin_layout Plain Layout
+
+
+\backslash
+hyperlink{kzsppom}{karakterizacijo zveznosti s pomočjo praslik odprtih množic}
+\end_layout
+
+\end_inset
+
+.
+ Dokazujemo torej,
+ da 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:g^{-1}\left(V\right)$
+\end_inset
+
+ je zopet odprta množica 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Velja 
+\begin_inset Formula $g^{-1}\left(V\right)=\left\{ x\in f\left(I\right);g\left(x\right)\in V\right\} =\left\{ x\in f\left(I\right):\exists v\in V\cap I\ni:x=f\left(v\right)\right\} =f\left(V\cap I\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Torej dokazujemo 
+\begin_inset Formula $\forall V^{\text{odp.}}\subset\mathbb{R}:f\left(I\cap V\right)$
+\end_inset
+
+ je spet zopet odprta 
+\begin_inset Formula $\subseteq f\left(I\right)$
+\end_inset
+
+,
+ kar je ekvivalentno
 \begin_inset Formula 
 \[
-\lim_{x\searrow0}\frac{1}{1+e^{1/x}}=0,\lim_{x\nearrow0}\frac{1}{1+e^{1/x}}=1
+\forall y\in f\left(I\cap V\right)\exists\delta>0\ni:\left(y-\delta,y+\delta\right)\cap f\left(I\right)\subset f\left(I\cap V\right).
 \]
 
 \end_inset
 
-Toda 
-\begin_inset Formula $\exists\lim_{x\to a}f\left(x\right)\Leftrightarrow\exists\lim_{x\nearrow a}f\left(x\right)\wedge\exists\lim_{x\searrow a}f\left(x\right)\wedge\lim_{x\nearrow a}f\left(x\right)=\lim_{x\searrow a}f\left(x\right)$
+Pišimo 
+\begin_inset Formula $y=f\left(x\right),x\in I\cap V$
 \end_inset
 
 .
+ Privzemimo,
+ da 
+\begin_inset Formula $f$
+\end_inset
+
+ narašča (če pada,
+ ravnamo podobno).
+ Ker jer 
+\begin_inset Formula $ $
+\end_inset
+
+
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Enakomerna zveznost
 \end_layout
 
 \begin_layout Definition*
-Označimo 
-\begin_inset Formula $\lim_{x\searrow a}f\left(x\right)\eqqcolon f\left(a+0\right),\lim_{x\nearrow a}f\left(x\right)\eqqcolon f\left(a-0\right)$
+\begin_inset Formula $f:I\to\mathbb{R}$
 \end_inset
 
-.
- Če 
-\begin_inset Formula $\exists f\left(a+0\right)$
+ je enakomerno zvezna na 
+\begin_inset Formula $I$
 \end_inset
 
- in 
-\begin_inset Formula $\exists f\left(a-0\right)$
+,
+ če 
+\begin_inset Formula 
+\[
+\forall\varepsilon>0\exists\delta>0\forall x,y\in I:\left|x-y\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Note*
+Primerjajmo to z definicijo 
+\begin_inset Formula $f:I\to\mathbb{R}$
+\end_inset
+
+ je (nenujno enakomerno) zvezna na 
+\begin_inset Formula $I$
 \end_inset
 
 ,
- vendar 
-\begin_inset Formula $f\left(a+0\right)\not=f\left(a-0\right)$
+ če
+\begin_inset Formula 
+\[
+\forall\varepsilon>0,a\in I\exists\delta>0\forall x\in I:\left|x-a\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(y\right)\right|<\varepsilon.
+\]
+
+\end_inset
+
+Pri slednji definiciji je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisna od 
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+ in 
+\begin_inset Formula $a$
 \end_inset
 
 ,
- pravimo,
- da ima 
-\begin_inset Formula $f$
+ pri enakomerni zveznosti pa le od 
+\begin_inset Formula $\varepsilon$
 \end_inset
 
- v točki 
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $f\left(x\right)=\frac{1}{x}$
+\end_inset
+
+ ni enakomerno zvezna,
+ ker je 
+\begin_inset Formula $\delta$
+\end_inset
+
+ odvisen od 
 \begin_inset Formula $a$
 \end_inset
 
- 
-\begin_inset Quotes gld
+.
+ Če pri fiksnem 
+\begin_inset Formula $\varepsilon$
 \end_inset
 
-skok
-\begin_inset Quotes grd
+ pomaknemo tisto pozitivno točko,
+ v kateri preizkušamo zveznost,
+ bolj v levo,
+ bo na neki točki potreben ožji,
+ manjši 
+\begin_inset Formula $\delta$
 \end_inset
 
 .
 \end_layout
 
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
 \begin_layout Corollary*
-sssssssssss 
+sssssssssss
 \end_layout
 
 \begin_layout Corollary*