is there a mistake in this proof that union of finite nowhere dense sets is nowhere dense

Question

I think there is a mistake near the end. The union of interiors is a subset of the interior of unions and not the other way around. Is there still a way to do a proof by contradiction along similar lines even though this proof is incorrect.

Answer 1

You’re right about the error. The argument can, however, be completed along these general lines. Let

$$U_0=\operatorname{int}\operatorname{cl}\bigcup_{k=1}^nA_k=\operatorname{int}\bigcup_{k=1}^n\operatorname{cl}A_k\,,$$

and suppose that $U_0\ne\varnothing$. For $m=1,\ldots,n$ let $U_m=U_0\setminus\bigcup_{k=1}^m\operatorname{cl}A_k$. Then

$$\varnothing\ne U_0\supseteq U_1\supseteq\ldots\supseteq U_n=\varnothing\,,$$

so there is a smallest $m\le n$ such that $U_m=\varnothing$. Then $U_{m-1}$ is a non-empty open set, and

$$U_{m-1}\setminus\operatorname{cl}A_m=U_m=\varnothing\,,$$

so $U_{m-1}\subseteq\operatorname{cl}A_m$. But then $\operatorname{int}\operatorname{cl}A_m\supseteq U_m\ne\varnothing$, contradicting the assumption that $A_m$ is nowhere dense.