Is every proper subset of the real numbers bounded by the set of real numbers? Trying to see if I understand the definition of bounded. Are these true:


$(-1, 1)$ is bdd above by $\mathbb{R}$ (and below)
$(-1, 1)$ is bdd above by $(-1, 1]\, $,  and $1$ is the supremum of $(-1,1)$
$[-1, 1]$ is bdd above and below by itself
$(-1, 1)$ is not bdd above by $[-1,1)$
$\mathbb{R}^+$ is not bdd above by $\mathbb{R}$ but it is bounded below, with an infimum of $0$.


Any mistakes or other elementary cases I should consider?
I have up-voted the answer by Xander Henderson. But since the subject line of the question says "Is every proper subset of the real numbers bounded by the set of real numbers?" it seems like a good idea to add that the set $\mathbb Z$ of all integers is a proper subset of $\mathbb R$ that is not bounded either above or below within $\mathbb R.$ (And I wrote "within $\mathbb R,$" not "by $\mathbb R.$")
That is why the real number system is 'Complete'. 
In case of open interval, say (a,b) we can take the deleted neighbourhood of a and b. Each member of the neighbourhood system lies in the set R. But, we cannot 'fix' a upper bound or lower bound.
Sorry for giving a rather informal answer, but I hope it'll be of use.