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I am preparing for my topology test and I came across the following question:

Proof [0,1] is not compact for the lower limit topology on $\mathbb{R}$.

My approach:

Let $\mathcal{A}=\{[1/n,2)|n\in\mathbb N\}\cup\{0\}$ be an open cover of [0,1]. Clearly, we can not extract a finite subcover from $\mathcal{A}$ such that the finite subcover covers the whole of $[0,1]$. Is this proof correct? I couldn't solve myself if $\{0\}$ is open in $([0,1],\mathcal{T}_{lowerlimit})$.

Thanks in Advance!

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Your suggestion is incorrect, as $\{0\}$ is not an open set in the lower limit topology. It's a cover, without a finite subcover, but not an open cover.

The problem lies at $1$ here: the open sets $[0,1-\frac{1}{n}), n \in \mathbb{N}$ together with $[1,2)$ do cover $[0,1]$ and are open in the lower limit topology on $\mathbb{R}$ (if you like we can replace the last one by $[1,2) \cap [0,1] = \{1\}$, which is thus an open set in the subspace topology of $[0,1]$, or you an use the open subsets of the whole space, it's equivalent to show (non-) compactness).

One can show that $A \subseteq (\mathbb{R}, \mathcal{T}_{\textrm{lower limit}})$ is compact iff it is closed and well-ordered by $>$), which in particular implies that $A$ is at most countable.

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Take $$A_n=[0,1-\frac{1}{n})\cup[1,2).$$ Clearly

$$[0,1] \subset \bigcup_{n=1}^\infty A_n, $$ but you can't extract a finite subcover.

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