How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly? How to prove that $\mathbb{Q} \subset \mathbb{R}$ is not locally compact directly? That is, how to construct a cover of an arbitrary neighborhood (e.g. $[0, 1] \cap \mathbb{Q}$) that does not have a finite subcover?
 A: Choose irrational point $i$ inside $[0,1]$. Take the class $O = \{[0, q) \cup (r, 1] \quad: \quad(q,r) \in \mathbb Q^2 \cap(0,1)^2,\quad q < i < r\}$ as your covering. Now we have a cover for the rationals in $[0,1]$ without a finite subcover, showing that the space is not compact.
For the sake of rigor: 
$O$ is an open cover in the subspace topology relative to the compact set $[0,1]$. 
$\cup O = [0,1]- \{i\}$, showing that we do, in fact, cover the rationals in $[0,1]$.
Consider the union of any finite subcover: $\cup_{k=1}^N[0,q_k)\cup(r_k,1]$ where $(q_k,r_k) \in \mathbb Q^2 \cap(0,1)^2,\quad\mathrm{and}\quad q < i < r$. This set is equivalent to $[0,\mathrm{max}_k\{q_k\})\cup(\mathrm{min}_k\{r_k\},1]$, where $k$ is understood to range from $1$ to $N$.
We conclude that any finite subcover misses the points in the interval  $[\mathrm{max}_k\{q_k\},\mathrm{min}_k\{r_k\}]$
In general, choose any neighborhood in $\mathbb R$. We must be able to find a segment $(a,b)$ within the neighborhood. Repeat the argument using segment $(a,b)$ instead of $[0,1]$.
We conclude that any neighborhood of $\mathbb R$ has an open cover for which no subcover can cover the rationals in that neighborhood.
A: If $\mathbb Q$ were locally compact, then for every rational q we would be able to find a 'hood (neighborhood) $U_q$ of q in which every sequence  has a convergent subsequence. But if we take an irrational x in $U_q$ (which always exists, since the irrationals are dense in $\mathbb R $), then a sequence ${x_n}$ approximating $x$ will not have a convergent subsequence. (e.g., if the interval contains $\sqrt 2$ , there cannot be a sequence of rationals converging to $\sqrt 2$), because the rationals are not order-complete.
A: Let's use that if a space (here $[0,1]\cap\mathbb Q$) has an open infinite partition, then it is not compact since the partition is an open covering that has no subcovering at all .   
Choose an increasing  sequence of irrational numbers $0\lt a_1\lt \ldots \lt a_n\lt\ldots\lt \sqrt 2/2$ converging to $\sqrt 2/2$ and a decreasing sequence of irrational numbers $1\gt b_1\gt \ldots \gt b_n\gt\ldots\gt \sqrt 2/2$ also converging to $\sqrt 2/2$.
Then contemplate the open partition (where $[[x,y]]$ means $[x,y]\cap\mathbb Q$)
$$[[0,1 ]]= [[0,a_1 ]]\sqcup  [[a_1,a_2 ]] \sqcup \ldots \bigsqcup \; \;\; [[b_1,1 ]]\sqcup  [[b_2,b_1 ]] \ldots                $$
A: Let $X:= \mathbb Q \cap [0,1]$. I'll show that if $U \subset X$ then $U$ doesn't have compact closure. Since $X$ is hausdorff this shows that it's not locally compact. It suffices to show that for any epsilon an open ball in X doesn't have compact closure. Given $x \in X$ and some $\epsilon>0$ pick an irrational $y \in B(x,\epsilon)$. Set $a=x-\epsilon$ and $b=x+\epsilon$, then the following is an open cover of $B(x,\epsilon)$ with no finite subcover:
$$\mathcal U := \left\{ \left( \left(a,y-\frac{1}{n}\right) \cup \left(y+\frac{1}{n},b\right) \right) \cap \mathbb Q \mid n \in \mathbb N\right\}.$$
To see this note that $\mathcal U$ is a nested covering and that for any $n$ it doesn't cover $B(x,\epsilon)$.
A: Q is any set from  negative infinite up to positive infinity,thus it is not bounded from both side and also it is not closed set.here Q is not compact because it fail heines therom,       Heines therom stated as any set S is said to be compact if and only if the set is both bounded and closed.
