Bounded - Continuous Relation How to solve the following question?  $$$$
Suppose $f:A\subset\Bbb{R}^2\to\Bbb{R}$ continuous in the rectangle $A=\{(x,y)\in\Bbb{R}^2|\alpha\leq x\leq\beta;\alpha'\leq y\leq\beta'\}.$ Proof that $f$ is bounded in this rectangle. ($f$ to be bounded in $A$ means that exists some $M>0$ such that $|f(x,y)|\leq M$ in $A$.)
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(Sugestion: suppose, by absurd, that $f$ is not bounded in $A$. Then, exists $(x_1,y_1)$ in $A$ such that $|f(x_1,y_1)|>1$. Taking the middle point of each side, divide the rectangle $A$ in four equal rectangles; in one of them, labeled $A_2$, $f$ won't be bounded, thus will exist $(x_2,y_2)\in A_2$ such that $|f(x_2,y_2)>2|$ etc.)
 A: You continue in the suggested way, so you get will $$A=A_1 \supset A_2 \supset A_3 \supset \cdots \supset A_n \supset \cdots$$
and $(x_n, y_n)$ such that $(x_n, y_n) \in A_n$ and $|f(x_n, y_n)| > n$
For $n, m >k$ you have both $(x_n, y_n)$ and $(x_m, y_m)$ are in $A_k$(since $A_k \supset A_m $ and $A_k \supset A_n $), and they distance is smaller than the length of the diagonal line of $A_k$, i.e. 
$$d\left((x_n, y_n), (x_m, y_m)\right) \leq \dfrac{l}{2^{k-1}}, \forall m,n >k$$
where $l$ is the length of the diagonal line of $A$.
So $(x_n, y_n)$ is a Cauchy sequence in $A$ and it converges to one point $(x,y)$ in $A$. Since $f$ is continuous, we get $$|f(x,y)| = \lim_{n\to +
\infty}|f(x_n, y_n)| = +\infty $$
Since $f$ should be finite at every point, we get a contradiction.
A: This indeed gives you a Cauchy-sequence $(x_n,y_n)$ in $A$ with $f(x_n,y_n)>n$. 
The sequence converges however to some $(x,y)\in A$ and the continuity of $f$ then demands that $f(x_n,y_n)$ converges to $f(x,y)$. Contradiction found!

edit:
In general a continuous function 'sends' compact sets (such as the rectangle) to compact sets. So $f(A)$ is a compact subset of $\mathbb R$ wich means in this context the same as closed and bounded.
