If A is a Subset of B, then the closure of A is a Subset of the closure of B.

Question

Conditions

Hello, I am trying to prove that $A \subseteq B \implies CL(A) \subseteq CL(B)$. I know how to prove when you define closure of set E as the Intersection of all sets that are closed and contain E, but I just want to prove it through the conditions in the image above. Sorry if it may seem obvious, but I don't see it. Also, the conditions are the Kuratowski closure axioms.

Answer 1

Suppose $A \subseteq B$. We have, by property 3:

\begin{equation*} \overline{B}=\overline{A \cup (B-A)} = \overline{A} \cup \overline{A-B} \end{equation*}

Therefore, $\overline{A} \subseteq \overline{B}$.

Answer 2

$(X, \tau) $ be a topological space.

$A, B\in 2^X$ with $A\subseteq B$.

Claim : $cl(A) \subseteq cl(B) $

$A\subseteq B\subseteq cl(B) $

Hence, $cl(B) $ is a closed set containing $A$ and $cl(A) $ by definition is the smallest closed set containing $A$.

Hence, $cl(A) \subseteq cl(B) $.