Walter Rudin, Principles of Mathematical, Question 2.7b.
Let $A_1, A_2,…$ be subsets of a metric space. If $B = \cup_{i=1}^\infty A_i$, prove that $\cup_{i=1}^\infty \bar{A_i}\subset \bar{B} $, where $\bar{A}$ is the closure of $A$.
Most of the answer(s) provided online proves only inclusion, not proper inclusion (i.e. $\cup_{i=1}^\infty \bar{A_i}\subseteq \bar{B}$, and not $\cup_{i=1}^\infty \bar{A_i}\subset \bar{B}$). Could anyone explain why proper inclusion in this case follows immediately from inclusion?
Edit: not a duplicate of (Prove that $\bigcup^{\infty}_{k=1}\overline{A_k} \subset \bar{B}$ if $B=\bigcup^{\infty}_{k=1}A_k $), but a case of ambiguity with the notation – Rudin uses $\subset$ in place of $\subseteq$ (Thanks Alberto!)
Best Answer
$\subset$ just means "subset of" in Rudin, not "proper subset of". He does not use $\subseteq$. See page 3 in chapter 1 where he introduces this notation.