Editing Tietze extension theorem (section)

==Proof==
The function <math>F</math> is constructed iteratively. Firstly, we define
<math display=block>
\begin{align}
c_0 &= \sup \{|f(a)|:a\in A\}\\
E_0 &= \{a\in A:f(a)\geq c_0/3\}\\
F_0 &=\{a\in A:f(a)\leq -c_0/3\}.
\end{align}
</math>
Observe that <math>E_0</math> and <math>F_0</math> are [[closed set | closed]] and [[disjoint sets | disjoint]] subsets of <math>A</math>. By taking a linear combination of the function obtained from the proof of [[Urysohn's lemma]], there exists a [[continuous function]]  <math>g_0:X\to \mathbb{R}</math> such that
<math display=block>
\begin{align}
	g_0 &= \frac{c_0}{3}\text{ on }E_0\\
	g_0 &= -\frac{c_0}{3}\text{ on }F_0
\end{align}
</math>
and furthermore 
<math display=block>-\frac{c_0}{3}\leq g_0 \leq \frac{c_0}{3}</math>
on <math>X</math>. In particular, it follows that 
<math display=block>
\begin{align}
	|g_0| &\leq \frac{c_0}{3}\\
	|f-g_0| &\leq \frac{2c_0}{3}
\end{align}</math>
on <math>A</math>. We now use [[mathematical induction | induction]] to construct a sequence of continuous functions <math>(g_n)_{n=0}^\infty</math> such that 
<math display=block>
\begin{align}
|g_n|&\leq \frac{2^nc_0}{3^{n+1}}\\
|f-g_0-...-g_{n}|&\leq \frac{2^{n+1}c_0}{3^{n+1}}.
\end{align}</math>
We've shown that this holds for <math>n=0</math> and assume that <math>g_0,...,g_{n-1}</math> have been constructed. Define
<math display=block>c_{n-1} = \sup\{|f(a)-g_0(a)-...-g_{n-1}(a)|:a\in A\}</math>
and repeat the above argument replacing <math>c_0</math> with <math>c_{n-1}</math> and replacing <math>f</math> with <math>f-g_0-...-g_{n-1}</math>. Then we find that there exists a continuous function <math>g_n:X\to \mathbb{R}</math> such that 
<math display=block>
\begin{align}
|g_n|&\leq \frac{c_{n-1}}{3}\\
|f-g_0-...-g_n|&\leq \frac{2c_{n-1}}{3}.
\end{align}
</math>
By the inductive hypothesis, <math>c_{n-1}\leq 2^nc_0/3^n</math> hence we obtain the required identities and the induction is complete. Now, we define a continuous function <math>F_n:X\to \mathbb{R}</math> as 
<math display=block>F_n = g_0+...+g_n.</math>
Given <math>n\geq m</math>,
<math display=block> 
\begin{align}
|F_n - F_m| &= |g_{m+1}+...+g_n|\\
&\leq \left(\left(\frac{2}{3}\right)^{m+1}+...+\left(\frac{2}{3}\right)^{n}\right)\frac{c_0}{3}\\
&\leq \left(\frac{2}{3}\right)^{m+1}c_0.
\end{align}
</math>
Therefore, the sequence <math>(F_n)_{n=0}^\infty</math> is [[Cauchy sequence | Cauchy]]. Since the [[metric space | space]] of continuous functions on <math>X</math> together with the [[uniform norm | sup norm]] is a [[complete metric space]], it follows that there exists a continuous function <math>F:X\to \mathbb{R}</math> such that <math>F_n</math> [[uniform convergence | converges uniformly]] to <math>F</math>. Since 
<math display=block>|f-F_n|\leq \frac{2^{n}c_0}{3^{n+1}}</math>
on <math>A</math>, it follows that <math>F=f</math> on <math>A</math>. Finally, we observe that 
<math display=block> 
|F_n|\leq \sum_{n=0}^\infty |g_n|\leq c_0
</math>
hence <math>F</math> is bounded and has the same bound as <math>f</math>. <math>\square</math>