Show that $\{[-n,n] | n \in \mathbb{N} \}$ is a basic cover of $\mathbb{R}$ Definition: Let $\mathcal{A}$ be a cover of a topological space $X$. The cover $\mathcal{A}$ is said to be a "basic cover" of $X$ if $U \subset X$ is open iff $U \cap A$ is open in $A$ for every $A \in \mathcal{A}$.
I tried to prove this: The family of subsets $\{[-n,n] | n \in \mathbb{N} \}$ is a basic cover of $\mathbb{R}$.
Proof:
Clearly $\{[-n,n] | n \in \mathbb{N} \}$ is a cover of $\mathbb{R}$ so I have to prove that if $U \cap [-n,n]$ is open in $[-n,n]$ for every $n \in \mathbb{N}$, then $U$ is open. I will show that $U$ is a neighborhood of every $x \in U$.
Let $x \in U$, $m:= \lfloor{|x|}\rfloor+1$; 
(If $y \in \mathbb{R}$, $\lfloor y \rfloor=\max\, \{k\in\mathbb{Z} : k\le y\}$).
Then $x \in (-m,m)$ and, since $U \cap [-m,m]$ is open in $[-m,m]$, there are four options:
$x \in U \cap [-m,m]= \begin{cases} (a,b) \\ [-m,m] \\ [-m,b) \\ (a,m] \end{cases} \quad $ with $\quad -m \le a < b \le m$
For each option I can choose an open set $V$ s.t. $x \in V \subset U$:
$V= \begin{cases} (a,b) \\ (-m,m) \\ (-m,b) \\ (a,m) \end{cases} \quad $ with $\quad -m \le a < b \le m$
Then $U$ is open.
 A: The '4 options' part is not clear, as not only intervals are open sets. 
But we can fix it by saying: since $U\cap [-m,m]$ is open in $[-m,m]$, there is an interval $I$ open inside $[-m,m]$, such that $x\in I\subseteq U$. 
For this interval $I$ we have the four possibilities... 
...and then we're done.
A: We prove the following:
Let $U\subseteq\mathbb{R}$. Suppose that for each $n\in\mathbb{N}$,
$U\cap[-n,n]$ is an open subset of $[-n,n]$ with respect to the
relative topology, then $U$ is an open subset of $\mathbb{R}$.
Proof: Let $x\in U$ be aribrary. Choose $n\in\mathbb{N}$ such that
$n>|x|+1$. Note that $x\in U\cap[-n,n]$ is an interior point of
$U\cap[-n,n]$ with respect to the relative topology of $[-n,n]$,
so there exists $\delta_{1}>0$ such that $B(x,\delta_{1};n)\subseteq U\cap[-n,n]$,
where $B(x,\delta_{1};n)=\{y\in[-n,n]\mid|y-x|<\delta_{1}\}$. Choose
$\delta_{2}=\min(\frac{1}{2},\delta_{1})>0$, we clearly have $x\in B(x,\delta_{2};n)\subseteq B(x,\delta_{1};n)\subseteq U\cap[-n,n]$.
Observe that $B(x,\delta_{2};n)=B(x,\delta_{2}):=\{y\in\mathbb{R}\mid|y-x|<\delta_{2}\}$.
For, obviously LHS is a subset of RHS. On the other hand, let $y\in\mbox{RHS}$,
then 
$$
|y|=|(y-x)+x|\leq|y-x|+|x|<\delta_{2}+|x|<n
$$
and hence $y\in[-n,n]$. Therefore $y\in\mbox{LHS}.$
We arrive the step: $x\in B(x,\delta_{2})\subseteq U\cap[-n,n]\subseteq U$.
Hence $x$ in an interior point of $U$ and this shows that $U$ is
an open subset of $\mathbb{R}$.
