Can anybody help me to prove the following result?
Corollary of Baire theorem: Let $(K_j)_{j>0}$ be an increasing sequence of compact sets in $C^n$ and $X$ a bounded open set such that $\overline X\subset\cup K_j.$ Then for every $x\in X$ there is $J>0$ so that $x\in int(K_J).$
Can anybody check whether this proof answers correctly the problem above?
For $j>0$ set $C_j=\mathbb{C}^m\setminus Int(\bigcup_{n=1}^j K_n)$ note that $C_j$ is closed and contains $\mathbb{C}^m\setminus K_j$ (because $K_j=\bigcup_n^j K_{n}$ since the sequence is increasing).
Let $x\in X$ we will prove by contradiction that $x\in Int(K_{j_0})$ for some $j_0>0.$ Suppose that $K_j$ does not contain any open ball centered at $x$, for any $j.$ This means that any open ball $B$ centered at $x,$ contains a point from $\mathbb{C}^m\setminus K_j$ in other word for all $j>0$ and $r>0$ the set $B(x,r)\setminus K_j$ is non-empty. That means for all $j>0$ there is $x_j\in\mathbb{C}^m$ so that $x_j\in B(x, 1/j)\setminus K_j.$
Remark that the sequence $(x_j)_j$ converges to $x$ and all its subsequences also converge to $x.$ As the sequence $(K_j)_j$ is increasing then for $j$ fix the point $x_n$ does not belong to $K_j$ for all $n\geq j$ that means $x_n\in C_j$ for all $n\geq j$ by closedness, the limit $x$ of the subsequence $(x_n)_{n>j}$ belongs to $C_j$ this for all $j>0$ in other word $x\in \cap^{\infty}_{j=1} C_j.$ That means $x\not\in Int(\bigcup^{\infty}_{n=1} K_n)$ then $x\not\in X,$ a contradiction hence there is $n_0>0$ so that $B(x, 1/n_0)\subset K_{n_0}$ that gives $x\in Int(K_{n_0})$.