Union/intersection over index families of intervals I am currently reading "Introduction to Topology" by Bert Mendelson (third edition).
At Chapter 1, Section 4 - Indexed Families of Sets.  For Exercise 5, he asks the following:

Let $I$ be the set of real numbers that are greater than $0$.
For each $x \in I$, let $A_x$ be the open interval $(0,x)$.
a) Prove that $\cap _{x \in I}\ A_x = \emptyset$
b) Prove that $\cup _{x \in I}\ A_x = I$.
For each $x \in I$, let $B_x$ be the closed interval $[0,x]$.
c) Prove that $\cap _{x \in I}\ B_x = \{0\}$
d) Prove that $\cup _{x \in I}\ B_x = I \cup \{0\}$.

Maybe I have misread it, but some of this doesn't make sense to me.
I am fine with Part a as for the smallest $x \in I$, the open interval $(0, x) = \emptyset$ given that $I$ is the set of real numbers greater than $0$.
Part c seems wrong: for each $x \in I$, the set $B_x$ should contain both $0$ and the smallest $x \in I$ - because $B_x$ is the closed interval $[0, x]$, where $0 \in [0, x]$ and $x \in [0, x]$.  As $\cap _{x \in I}\ B_x$ will contain all common elements of $B_x$ (for each $x \in I$), it seems to me that it should be: $\cap _{x \in I}\ B_x = \{0, smallest(I)\}$.
As I was writing this question I originally had trouble with b and d too, however I have since changed my mind - as I was forgetting that $I$ is an infinite set.  I was thinking up issues when considering the largest $x \in I$.
So my question is: am I mistaken with part c? and how so?

There is no smallest member of $I$. Given any $x \in I$, you'll eventually come across $[0, \frac{x}{2}]$ in your intersection. The only common point to all those intervals is $0$.
