Prove that $[0,1]$ is not a compact subset of $\mathbb{R}$ with the lower limit topology, i.e. open sets are of the form $[a,b)$. Prove that $[0,1]$ is not a compact subset of $\mathbb{R}$ with the lower limit topology, i.e. open sets are of the form $[a,b)$.
My question is will different topology affect compactness of a set? If this is so, why? 
At first, when I see this question, I thought something is wrong with this question because I know that $[0,1]$ is compact by using Heine-Borel theorem. 
 A: It is not compact : Take the open cover
$$
\{(-\infty, 1-1/n)\}_{n=1}^{\infty} \cup \{[1,\infty)\}
$$
In fact, compact sets in this topology are necessarily countable.
So yes, compactness certainly depends on the topology.
A: Two answers have already demonstrated that the interval is not compact, by exhibiting simple open covers for it with no finite subcover.
The other half of the question is why compactness depends on the topology.  I think in this case it might be instructive to consider the discrete topology.  In the discrete topology, no infinite set $S$ is compact, because one can exhibit the open cover consisting of the sets $\{s\}$ for each $s\in S$. Since each point of $S$ is contained in exactly one of the elements of the cover, none of the elements can be omitted, and the cover not only has no finite subcover, it has no proper subcover at all!  So in the discrete topology, a set is compact if and only if it is finite, which is quite different from the situation in a metric space.
The situation in the indiscrete topology is different in exactly the opposite way;  every set is compact. The only nonempty set is the entire space, so there are no infinite open covers to begin with; every open cover consists of just this one open set, and a finite cover obviously has a finite subcover. 
A: In that case you can't use the Heine-Borel theorem because this theorem only apply to the case of normal topology.
To prove this proposition, you just find the open cover of $[0,1]$ such that every finite subcover does not cover $[0,1]$. Let $\mathcal{A}$ be a set of open sets defined as
$$\mathcal{A}=\{[0,r):0<r<1\}\cup \{[1,2)\}$$
then $\mathcal{A}$ cover $[0,1]$. However every finite subcover of $\mathcal{A}$ does not cover $[0,1]$.
A: GRE9367 #62



Ian Coley's solution:



Sean Sovine's solution:




To prove $X$ is not compact, my first proof was similar to Ian Coley's, but I came up with another proof:

If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $\mathbb R$. This implies that the topologies are equal by (*), a contradiction.


(*) Munkres Exer26.1 (dbfin pf)



