Let \(\displaystyle \mathcal{F}=B_{n}, \;\ \text{where} \;\ B_{n}=(0,3-\frac{1}{n}), \text{for} \;\ n\in \mathbb{N}\), Then, \(\displaystyle \mathcal{F}\) is an open cover for [1,3), which has no finite subcover.
Let \(\displaystyle \mathcal{F}=B_{n}, \;\ B_{n}=(-n,n)\). Then, \(\displaystyle \mathcal{F}\) is an open cover of \(\displaystyle \mathbb{N}\) which has no finite subcover.
Think along the same lines as the other two. Let \(\displaystyle \mathcal{F}=B_{n}\), where \(\displaystyle B_{n}=(\frac{1}{n},2), \;\ \text{for} \;\ n\in \mathbb{N}\)
Then \(\displaystyle \mathcal{F}\) is an open cover of \(\displaystyle \mathbb{N}\) which has no finite subcover.
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