When are supremum/infimum of sets values not contained in this set?

Question

I am a bit unsure about how the supremum/infimum of a set is obtained, in particular in the following simple toy example.

Let $f\colon \mathbb{R}\rightarrow \mathbb{R}$ be the identity function, i.e. $f(x)=x$ for all $x$.

Now I am interested in $a:=\inf\{x \colon f(x) > 0 \}$ and $b:=\inf\{x > 0 \colon f(x) > 0\}$.

I wonder whether $a=b$ and whether $a$ or $b$ are equal to 0.

We had in a lecture about probability theory the following examples:

In one example we considered for the distribution function $F$ and a $x \in [0,1]$ the quantity

$\sup\{y \in \mathbb{R} \colon F(y) < x\}$.

The picture drawn corresponding to this example suggests that when the function is continuous at the point $y^*$ such that $F(y^*)=x$ and $y^*$ is the smallest value mapped to $x$, then the supremum is $y^*$, even though $y^*$ does not satisfy the property that $F(y^*) < x$.

In some other example we had $\sup A_n = \emptyset$ with $A_n:=(0,1/n)$.

Now I wonder why it is not the case that $\sup A_n = 0$, as in the above example we also picked a number not part of the set, and here $0$ is not part of the set as well, but would satisfy the other property of being $<1/n$ for all $n$.

Answer 1

An example; On the real line open sets do not contain thier supremum.

If a set is open then you can always find smaller numbers then a given number you might claim being your sup and still stay outside the set. In a closed set the boundary possess a property that makes this impossible since any ball around a point on the boundary intersect both the set and the complement.

Answer 2

If $F$ is a (cumulative) distribution function, we have $\lim_{x \downarrow \infty} F(x) = 0$ and $\lim_{x \uparrow \infty} F(x) = 1$, and $F$ is non decreasing and continuous from the right.

It follows that the set $S = \{y | F(y) < x \}$ must either (1) be empty, (2) be the real line or (3) have the form $(-\infty,t)$ for some $t$ (in which case we have $\sup S = t$).

In Case (3), $S$ cannot include $t$ since $F$ is right continuous. We must have $F(t) \ge x$.

If $F$ is continuous then in Case (3) we must have $F(t) = x$. Furthermore, if $F(s) = x$ for some $s < t$, we must have $F(y) = x$ for $y \in [s,t]$ and so $[s,t] \cap S = \emptyset$ which contradicts $S$ having the form $(0,t)$. Hence, when $F$ is continuous, we see that $\sup S =t= \inf \{ s | F(s) = x \}$, and we must have $F(t) = x$ and $F(s) < x$ for $s <t$.