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Assuming I choose a rectangle from the form: $$I=[a_1,b_1]\times \cdots \times [a_n,b_n]=\{x\in \mathbb R^n |a_i\leq x_i\leq b_i for 1\leq i\leq n\}$$

Can it be said that I chose it to be closed and bounded? If so, how can I show that?

havakok
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    See here: http://math.stackexchange.com/questions/567335/cartesian-product-of-compact-sets-is-compact – smcc Aug 06 '16 at 11:15

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If you say that, you mean, that either $I = \emptyset $ (which happens if $b_i < a_i$ for some $i$), or $I \ne \emptyset$ and all $a_i, b_i$ are finite.

I would assume that anyone who understands "closed" and "bounded" would know that a parallelepiped as $I$ is closed and bounded, and there is nothing you hae to show.

At some point, for your own benefit, you should check $I$ is closed and bounded. But you have to do it only once. To show that $I$ is closed, you can show that is a sequence of points $H_h \in I$ converges to a limit $H$, thenthe limit point is in $I$, which can be done using basic properties of limits. Use that all boundary points of the intervals are finite to show $I$ is bounded.

VictorZurkowski
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