Given intervals $I_n = (1 - 1/n, 3 - 1/n]$ for all natural numbers n,
I have been told that the intersection of all these intervals is $[1, 3)$
However, I cannot figure out why.
I would expect 1 to not be in the intersection since it is on the open side of the intervals and 3 to remain in the intersection.
However, the opposite has happened. 1 is in the set and 3 is out.
Would be greatly appreciated it if someone could explain to me how this happened.
You've been told wrong, since $I_1 = (0,2]$ then $3 \not \in I_1$ and so it is not in the intersection. If you meant to type $3 + \frac{1}{n}$ this is still incorrect as $\bigcap I_n = [1,3]$. For $x$ to be in the intersection we must have that $x \in I_n$ for all $n$.
Remember that $\left(1 - \frac{1}{n}, 3 + \frac{1}{n}\right]$ is the set $\{x \in \mathbb{R} : 1 - \frac{1}{n} < x \leq 3 + \frac{1}{n}\}$ and since $1 - \frac{1}{n} < 1$ for all $n$ and $3 \leq 3 + \frac{1}{n}$ for all $n$ we have that $[1,3] \subset \bigcap I_n$.
I'll leave it to you to figure out why for $x < 1$ or $x > 3$ we have that $x \not \in I_n$ for some $n$. This means the intersection is exactly $[1,3]$.