Increasing sequence of compacts of $(0,1)$

Question

I first recall a definition:

Let $\Omega$ an open subset of $\mathbb{R}^n$. A sequence $(K_n)$ of compact subsets of $\Omega$ is said to be exhaustive if we have: $$\forall n\in\mathbb{N},\,K_n\subset\mathring{K_{n+1}}$$ and $$\bigcup_{n\in\mathbb{N}}K_n=\Omega$$ We can show that for each open subset $\Omega\subset\mathbb{R}^n$, there exist exhaustive sequences of compacts, and that if $(K_n)$ is such a sequence, then for any compact $K\subset\Omega$, $K$ is a subset of one of the $K_n$.

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Now here's my problem:

I take $n=1$ and $\Omega=(0,1)$. I want to find an increasing sequence of compacts $(K_n)$ whose union is $(0,1)$ and a compact $K\subset (0,1)$ such that for any $n\in\mathbb{N}$, $K\not\subset K_n$.

$(K_n)$ is increasing so we have $K_n\subset K_{n+1}$ for all $n$, but we can't have $K_n\subset\mathring{K_{n+1}}$ for all $n$, otherwise the sequence would be exhaustive and $K$ wouldn't exist. So I think there is something to do with the boundaries of the compacts, but I can't figure out what.

The first thing I thought of was $$K_n=\left[\frac{1}{n},1-\frac{1}{n}\right]$$ but it's exhaustive...

Answer 1

For simplicity of the construction, consider $(-1,1)$.

Let $K = \{0\} \cup \{\frac1{n+1}: n \in \Bbb N\}$.

Define $K_n$ as: $$K_n = \left[-\frac{n}{n+1},\frac{n}{n+1}\right] \setminus \bigcup_{k = n}^\infty \left(\frac1{4k+3},\frac1{4k+1}\right)$$

Then it is manifest that $K_n$ is increasing and its union is $(-1,1)$. However, for each $n$, $\dfrac1{4n+2} \in K$, but $\dfrac1{4n+2} \notin K_n$. Hence $K \not\subset K_n$.

Answer 2

Next try ;-)

Choose $0< a < b < c < d$ and let $K_1 = [a,b]\cup[c,d]$

Choose exhaustive sequnces $L_i, M_i, N_i$ of $(0, a), (b,c)$ and $(d,1)$ respectively, then let $K_i= K_1 \cup L_i \cup M_i \cup N_i$ The union of the $K_i$ will cover $(0,1)$, but the interval $[b,c]$ will not be contained in any finite $K_i,$ because otherwise $(b,c)$ would be contained in a finite number of the $M_i$.