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Let $(X,\mathfrak T)$ be a topological space and suppose that A and B are subsets of X such that $A \subseteq B$ then $Cl(A) \subseteq Cl(B)$.

My definition of closure is "Let $(X, \mathfrak T)$ be a topological space and let $A \subseteq X$. The closure of $A$ is defined by $ Cl(A) =\bigcap \{U \subseteq X : U$ is a closed set and $A \subseteq U\}$. From the definition I know that $A \subseteq Cl(A)$.

Here is my proof: Let $ x \in Cl(A)$ by the definition of closure $A \subseteq Cl(A)$ therefore $x \in A$. Since $A \subseteq B$. $ x \in B$ then $x \in Cl(B)$ since $B \subseteq Cl(B)$ therefore $Cl(A) \subseteq Cl(B)$.

How does my proof look? All of the proofs on topologies I have been doing involve set theory which is why I started with an element of one set and tried to show it was an element in the other set.

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This is not correct. The very first thing you say is that $x\in Cl(A)$, $A\subseteq Cl(A)$ implies $x\in A$... and it need not be. The closure of $A$ is contains $A$ itself, but may contain points not in the original set $A$.

Here, you need to use the definition of the closure: namely, that if $A\subseteq U$ and $U$ is closed, then $Cl(A)\subseteq U$ as well. (You can, and should, think of the closure as the smallest closed set that contains $A$.)

Now, consider $Cl(B)$. This is a closed set (it is the intersection of a collection of closed sets). Further, we know that $A\subseteq B$ and $B\subseteq Cl(B)$, so that $A\subseteq Cl(B)$.

So, $Cl(B)$ is a closed set that contains $A$, and therefore $Cl(A)\subseteq Cl(B)$, as desired.

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  • $\begingroup$ So I should start by letting x be in A? $\endgroup$
    – user219081
    Commented May 2, 2015 at 20:19

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