How to show $\bigcap_{n=1}^{\infty}[0,1+1/n) = [0,1]$? To show that they are equal, I need to show
$\bigcap_{n=1}^{\infty}[0,1+1/n) \subset [0,1]$ and $[0,1] \subset \bigcap_{n=1}^{\infty}[0,1+1/n)$
My attempt is: let $x \in [0,1] \Rightarrow 0 \leq x \leq 1$, since $1 < 1+1/n, \ \forall n \geq1 \Rightarrow x \in \bigcap_{n=1}^{\infty}[0,1+1/n) \Rightarrow [0,1] \subset \bigcap_{n=1}^{\infty}[0,1+1/n)$
However, I don't know how to show $\bigcap_{n=1}^{\infty}[0,1+1/n) \subset [0,1]$. It seems obvious since $\lim_{n \to \infty}  1+1/n = 1$, but I am having trouble to proving that. Any help or hint would be appreciated
 A: Observe that if $x \in \bigcap_{n=1}^{\infty}[0,1+1/n)$  then you have that
$$x \in [0,1+1/n), \forall  n \in \mathbb{N}\hspace{3mm} (1)$$
That implies that $x \geq 0$ and it remains to prove that $x \leq 1$.
Here you have two options:

*

*If you have studied about sequence limits, defining $x_n=1+1/n$ we get from (1)  that
$$x \leq x_n, \forall  n \in \mathbb{N}$$
So, by the monotonicity of the limit and knowing that $x_n \to 1$ we get that $x \leq 1$ as wanted.

*The second option is to prove it from scratch. Suppose by way of contradiction that $x >1$, then $x-1>0$ and there exists an $n_0 \in \mathbb{N}$ such that $x-1 > \frac{1}{n_0}$ $\left(\text{just take }  n_0 > \frac{1}{x-1} \right)$. Now, this implies that $x>1+1/n_0$ so $x \not \in [0,1+1/n_0)$ in contradiction with (1).

A: To simplify this, let $p$ be the statement $x\in [0,1]$ and $q$ be the statement $x\in \bigcap_{n=1}^{\infty}[0,1+\frac{1}{n})$
$x\in [0,1]\subset x\in \bigcap_{n=1}^{\infty}[0,1+\frac{1}{n}) \text{ is the same as } p\to q$
and $x\in \bigcap_{n=1}^{\infty}[0,1+\frac{1}{n})\subset x\in [0,1] \text{ is the same as } q\to p \text{ which is equivalent to } \lnot p \to \lnot q$.
As such, since you have proven that $x\in [0,1]\subset x\in \bigcap_{n=1}^{\infty}[0,1+\frac{1}{n})$ ($p\to q$, though may need to be more rigorous especially if for a class). For the next part it may help to assume that $x\notin [0,1]$ and try to conclude that $x\notin \bigcap_{n=1}^{\infty}[0,1+\frac{1}{n})$.
A: Showing
$$[0,1] \subset \bigcap_{n=1}^{\infty}[0,1+1/n)$$
is equivalent to showing
$$ \overline{[0,1]} \supset \overline{\bigcap_{n=1}^{\infty}[0,1+1/n)}. \tag1$$
Since
$$ \overline{[0,1]}=(-\infty, 0) \cup (1, \infty)$$
and
$$\overline{\bigcap_{n=1}^{\infty}[0,1+1/n)}=\bigcup_{n=1}^{\infty}(-\infty,0)\cup[1+1/n,\infty)=(-\infty,0)\cup\bigcup_{n=1}^{\infty}[1+1/n,\infty), $$
(1) becomes
$$ (-\infty, 0) \cup (1, \infty) \supset (-\infty,0)\cup\bigcup_{n=1}^{\infty}[1+1/n,\infty). \tag2$$
Now we show
$$ \bigcup_{n=1}^{\infty}[1+1/n,\infty)\subset(1,\infty). \tag3 $$
In fact, for $\forall x\in\bigcup_{n=1}^{\infty}[1+1/n,\infty)$, then $\exists n$ such that $x\in[1+1/n,\infty)$ and so $x\in(1,\infty)$; namely, (3) is true. Thus
$$ (-\infty,0)\cup\bigcup_{n=1}^{\infty}[1+1/n,\infty)\subset(-\infty, 0) \cup (1, \infty) $$
or (2) is true.
A: If $x$ is in the intersection, that means that $0\leq x<1+\frac{1}{n}$ for all $n\geq1$, using the fact that if $w_n<u_n<v_n$ for all $n$, then $\lim_n w_n\leq \lim_n u_n \leq \lim_n v_n$,  and since $\lim_n 0=0$ and $\lim_n x=x$ (constante sequences) then $0\leq x\leq 1$.
