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\centerline{\bf Sample Proof:  By Dr. Sarah}
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\begin{prob}
Prove that the standard topology $(\mathbb{R}^2,d)$ equals the square topology $(\mathbb{R}^2,\rho)$.
\end{prob}

\begin{proof}[Proof:]
To show that  $(\mathbb{R}^2,d) = (\mathbb{R}^2,\rho)$, we will show that $(\mathbb{R}^2,d) \subseteq (\mathbb{R}^2,\rho)$ and 
$(\mathbb{R}^2,d) \supseteq (\mathbb{R}^2,\rho)$, by showing that an open set in each metric is open in the other.

 $\supseteq$  
 Let $U \in  (\mathbb{R}^2,\rho)$ and let $x \in U$.  We will show that $U$ is open in  $(\mathbb{R}^2,d)$, ie that
 $\exists \epsilon >0$ so that $B_d(x,\epsilon) \subseteq U$.    By the definition
of open in $ (\mathbb{R}^2,\rho)$ we know $$(1)\hspace*{.2cm}  \exists \epsilon >0 \mbox{ so that } 
B_\rho(x,\epsilon) \subseteq U.$$  Take the same $\epsilon$ and let 
$ y \in B_d(x,\epsilon)$.   We will show that $ y \in U$ by first showing that  $ B_d(x,\epsilon) \subseteq B_\rho(x,\epsilon)$.  By definition of metric balls in the standard topology, we know that $d(x,y) < \epsilon$.  In addition, by a problem on the Metric Space Exercises (Mendelson p. 34 \# 3),
we have already proved that $\rho(x,y) \leq d(x,y)$, and so $\rho(x,y)  < \epsilon$.   Hence 
 $ y \in B_\rho(x,\epsilon)$.  In addition, 
by (1) we have $B_\rho(x,\epsilon) \subseteq U$.
Therefore $y \in B_d(x,\epsilon) \subseteq B_\rho(x,\epsilon) \subseteq U$, as desired.  

$\subseteq$
Let $U \in (\mathbb{R}^2,d)$ and let $x \in U$.  We will show that $U$ is open in  $(\mathbb{R}^2,\rho)$,  ie that
 $\exists \epsilon' >0$ so that $B_\rho(x,\epsilon') \subseteq U$.  
 By the definition
of open in $ (\mathbb{R}^2,d)$ we know $\exists \epsilon >0$ so that 
$B_d(x,\epsilon) \subseteq U$.  Take $\epsilon'=\frac{\epsilon}{\sqrt{2}}$. Let 
$y \in B_\rho(x,\epsilon') = B_\rho(x,\frac{\epsilon}{\sqrt{2}})$.  To show that $y \in U$, first we show 
that $y \in B_d(x,\epsilon)$ using 
the same Mendelson problem as in the  $\supseteq$ portion of the proof.  From this prior work, 
$d(x,y) \leq \sqrt{2} \rho(x,y) < \sqrt{2} \frac{\epsilon}{\sqrt{2}} = \epsilon$.
Hence $y \in B_d(x,\epsilon).  $  In addition, by our original choice of $\epsilon$ in the definition of open in 
 $ (\mathbb{R}^2,d), B_d(x,\epsilon) \subseteq U$.
 Thus  $B_\rho(x,\epsilon')=B_\rho(x,\frac{\epsilon}{\sqrt{2}})\subseteq 
B_d(x,\epsilon) \subseteq U$  so U is open in
 $(\mathbb{R}^2,\rho)$.

\end{proof}




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