View source for Theorem:Bolzano-Weierstrass

===Theorem (Bolzano&mdash;Weierstrass)===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">Every bounded sequence of real numbers contains a convergent subsequence.</blockquote>

Let (''x''<sub>''n''</sub>) be a sequence of real numbers bounded by a real number ''M'', that is  |''x''<sub>''n''</sub>|&nbsp;&lt;&nbsp;''M'' for all ''n''.  We define the set ''A'' by ''A''&nbsp;=&nbsp;{''r''&nbsp;|&nbsp;|''r''|&nbsp;&le;&nbsp;''M'' and ''r''&nbsp;&lt;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n''}.  We note that ''A'' is non-empty since it contains &minus;''M'' and ''A'' is bounded above by ''M''.  Let ''x''&nbsp;=&nbsp;sup&nbsp;''A''.  

We claim that, for any ε&nbsp;&gt;&nbsp;0, there must be infinitely many points of ''x''<sub>''n''</sub> in the interval (''x''&nbsp;&minus;&nbsp;ε,&nbsp;''x''&nbsp;+&nbsp;ε).  Suppose not and fix an ε&nbsp;&gt;&nbsp;0 so that there are only finitely many values of ''x''<sub>''n''</sub> in the interval (''x''&nbsp;&minus;&nbsp;ε,&nbsp;''x''&nbsp;+&nbsp;ε). Either ''x''&nbsp;&le;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n'' or ''x''&nbsp;&le;&nbsp;''x''<sub>''n''</sub> for at most only finitely many ''n'' (possibly no ''n'' at all).  Suppose ''x''&lt;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n''.  Clearly in this case ''x''&nbsp;&ne;&nbsp;''M''.  If necessary restrict ε so that ''x''&nbsp;+&nbsp;ε&nbsp;&le;&nbsp;''M''.  Set ''r''&nbsp;=&nbsp;''x''&nbsp;+&nbsp;ε/2 we have that ''r''&nbsp;&lt;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n'' because there are only finitely many ''x''<sub>''n''</sub> in the set [''x'',''r''] and ''x'' must be less than infinitely many ''x''<sub>''n''</sub>, furthermore |''r''|&nbsp;&lt;&nbsp;''M''. Thus ''r'' is in ''A'', which contradicts that ''x'' is an upper bound for ''A''. Now suppose ''x''&lt;&nbsp;''x''<sub>''n''</sub> for at most finitely many ''n''. Set ''y''&nbsp;=&nbsp;''x''&nbsp;&minus;&nbsp;ε/2.  Then there are at most only finitely man ''n'' so that ''x''<sub>''n''</sub>&nbsp;&ge;&nbsp;''y''.  Thus, if ''r''&nbsp;&lt;&nbsp;x<sub>''n''</sub> for infinitely many ''n'', we have that ''r''&nbsp;&le;&nbsp;''y''.  This means that ''y'' is an upper bound for ''A'' that is less than ''x'', contradicting that ''x'' wast the ''least'' upper bound of ''A''.  In either case we arrive at a contradiction, thus we must have that for any ε&nbsp;&gt;&nbsp;0, there must be infinitely many points of ''x''<sub>''n''</sub> in the interval (''x''&nbsp;&minus;&nbsp;ε,&nbsp;''x''&nbsp;+&nbsp;ε).
    
Now we show there is a subsequence that converges to ''x''.  We define the subsequence inductively, choose any ''x''<sub>n<sub>1</sub></sub> from the interval (''x''&nbsp;&minus;&nbsp;1,&nbsp;''x''&nbsp;+&nbsp;1).  Assuming we have chosen ''x''<sub>''n''<sub>1</sub></sub>,&nbsp;…,&nbsp;''x''<sub>''n''<sub>''k''&minus;1</sub></sub>, choose ''x''<sub>''n''<sub>''k''</sub></sub> to be an element in the interval (''x''&nbsp;&minus;&nbsp;1/k,&nbsp;''x''&nbsp;+&nbsp;1/k) so that ''n''<sub>''k''</sub>&notin;{''n''<sub>1</sub>,&nbsp;…,&nbsp;''n''<sub>k&minus;1</sub>}, this is possible as there are infinitely many elements of (''x''<sub>''n''</sub>) in the interval. Notice that for this choice of ''x''<sub>''n''<sub>''k''</sub></sub> we have that |''x''&nbsp;&minus;&nbsp;''x''<sub>''n''<sub>''k''</sub></sub>|<1/k.  Hence for any ε>0, if we take any ''k''&nbsp;&gt;&nbsp;1/ε, then  |''x''<sub>''n''<sub>''k''</sub></sub>-x|&nbsp;&lt;&nbsp;ε.  That is the subsequence (''x''<sub>''n''<sub>k</sub></sub>)&nbsp;→&nbsp;''x''. <math>\Box</math>