Problem of the Week #119 - September 8th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Let $A\subset X$, $f:A\rightarrow Y$ be continuous, and $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g:\overline{A}\rightarrow Y$, then $g$ is uniquely determined by $f$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Ackbach and Euge. You can find Ackbach's solution below.

[sp]Let $h:\overline{A}\to Y$ be a continuous extension of $f$ to $\overline{A}$. That is, $f(x)=h(x) \; \forall x\in A$. By assumption, $f(x)=g(x) \; \forall x\in A$. Therefore, $g(x)=h(x)\; \forall x\in A$.

WTS: $g(x)=h(x) \; \forall x\in\overline{A}\setminus A$.

Let $\varepsilon>0$, and let $x\in\overline{A}\setminus A$ be arbitrary, and suppose $\{x_{i}\}\to x$ to be a sequence in $A$ converging to $x$. Now $g(x_i)=h(x_i) \; \forall i$, since $x_i\in A$. Since $g$ and $h$ are both continuous, it follows that $\{g(x_i)\}\to g(x)$ and $\{h(x_i)\}\to h(x)$. Then there exists $N_g$ s.t. $|g(x_n)-g(x)|<\dfrac{\varepsilon}{2} \; \forall n>N_g$, and there exists $N_h$ s.t. $|h(x_n)-h(x)|<\dfrac{\varepsilon}{2} \; \forall n>N_h$. Let $N=\max(N_g, N_h)$, and suppose $n>N$. Then
\begin{align*}
|g(x)-h(x)|&=|g(x)-g(x_n)+g(x_n)-h(x)| \\
&=|g(x)-g(x_n)+h(x_n)-h(x)| \\
&\le |g(x)-g(x_n)|+|h(x_n)-h(x)| \\
&\le \frac{\varepsilon}{2}+\frac{\varepsilon}{2} \\
&=\varepsilon.
\end{align*}
Since $\varepsilon$ was arbitrary, $|g(x)-h(x)|=0$, or $g(x)=h(x)$. Since $x$ was arbitrary, $g=h \; \forall x\in \overline{A}$, and we are done.[/sp]