Answer:
Step-by-step explanation:
1)
We can find an open interval [tex](-\frac{1}{n}, 1+\frac{1}{n})[/tex]. Which contains the closed interval [tex][0,1][/tex] when [tex]n \to \infinity[/tex]
Therefore, the intersection of all the open intervals containing [tex][0,1[/tex[ is [tex][0,1][/tex]
2)
We can find closed intervals containing [tex][0,1][/tex] in [tex](-\frac{1}{n}, 1+\frac{1}{n})[/tex] when [tex]n \to \infinity[/tex]
Since [tex](-\frac{1}{n}, 1+\frac{1}{n})[/tex] is closed, the intersection of all those intervals containing [tex][0,1[/tex[ is the closed interval [tex][0,1][/tex]