I'm working on understanding double & triple integrals (and 3-space geometry in general) and I often encounter in my textbooks the requirement that f is continuous on "a closed bounded domain". Wouldn't these concepts be the same thing? Like, if a domain is closed, it contains it's endpoints, and it thus necessarily finite, and if it is bounded it is contained within some "ball" of finite radius centered around the origin and is so finite. I can't really imagine a domain being closed, and not bound, or vice versa. Am I missing some detail in distinguishing these two?
$\endgroup$ 72 Answers
$\begingroup$Since you are working with double integrals, I'll make some proofs in the space $\mathbb{R}^2$
Proposition 1: If $X$ is closed then it is bounded.
That proposition is false.
Counter example: $X = \mathbb{R}^2 \subset \mathbb{R}^2$
The complement of $\mathbb{R}^2$ is $\emptyset$, which is an open set.
Hence, $\mathbb{R}^2$ is closed. (also, feel free to use any other characterization of a closed set)
However, for each disk $B(0,r)$ of radius $r$, there exists a $x \in \mathbb{R}^2$ such that $||x|| > r$.
Thus, $\mathbb{R}^2$ is not bounded.
Proposition 2: If $X$ is bounded then it is closed.
That proposition is false.
Counter example: $X = B(0,1) = \{x\in\mathbb{R}^2, ||x|| < 1\}$, the unit circle around the origin in $\mathbb{R}^2$.
$X$ is bounded because $X \subset B(0,2)$ (for every $x \in X$, $||x|| < 2$)
However, $X$ is not closed because the sequence $x_n = (\frac{n}{n+1}, 0) \in X$ converges to $x =(1,0)$, but $x = (1,0) \notin X$.
Proposition 3: If $X$ is closed then it is finite
That proposition is false.
Counter example: $X = \mathbb{R}^2 \subset \mathbb{R}^2$
$X = \mathbb{R}^2$ has been proven to be closed in the propositions above.
$X = \mathbb{R}^2$ is trivially infinite because the injection $f(n) = (n, 0), n \in \mathbb{N}$ exists.
Proposition 4: If $X$ is bounded then it is finite
That proposition is false.
Counter example: $X = \bar{B}(0,1) = \{x\in\mathbb{R}^2, ||x|| <= 1\}$, the closed unit circle around the origin in $\mathbb{R}^2$.
I am gonna leave the exercise of proving that the closed unit disk is a closed set. (I could not fit that proof in one sentence).
$X = \bar{B}(0,1)$ is not finite because the injection $f(n) = (\frac{1}{n}, 0), n \in \mathbb{N}$ exists.
Okay, now I can give some intuition.
Please, take the following statements with caution since they are pretty informal.
Think of closed sets as sets that have bounds. A closed disk is closed. The upper plane including the line that divides it with the lower plane is closed.
Think of bounded sets as sets that can be put inside a disk. So, things that once you "zoom out enough" you will eventually be able to see the entire set inside a disk.
There are sets that are both bounded and closed (such as the closed unit disk), but one thing does not imply the other. Also, be careful so you do not mix those 2 concepts with the concept of being finite or infinite.
In a bounded set, the endpoints need not necessarily be a part of the set whereas in a closed set, the endpoints need to be a part of that set (as you have mentioned in your question). E.g. [0,1] and [0,1) are both bounded (by 0 and 1), but the second set isn't closed.
$\endgroup$
