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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 $Int(A) \subseteq Int(B)$.

I know this a true statement so now I need to go about proving it. I am in an introduction to proofs class and we are learning set theory and topology as a basis for prove writing. I only have a basic knowledge of topology and typically our proofs involve the definition and sets/elements.

My definition for interior is: Let $(X, \mathfrak T)$ be topological space and let $A \subset X$. The interior of $A$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.

Based on the definition I know that $Int (A) \subseteq A$. Here is my proof: Let $x \in Int(A)$ then $x \in (A)$ since $A \subseteq B$ then $x \in B$.. I feel like I am on the right track however I know that just because an element is in A does not necessary mean that it is in the interior of A since A is an open set.

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Your proof only shows interior point of A is in $B$, but doesn't show it's interior point of $B$.

Let $x\in int(A)$. Then by definition, there exists an open set $U$ such that $x \in U \subseteq A$.Sinc $A\subset B$, hence $x \in U \subseteq B$, then by definition, $x\in int(B)$.

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  • $\begingroup$ I wasn't sure if I could make that step as I was worried that U was an open set but I think I am still getting open set and open interval confuse me. $\endgroup$
    – user219081
    Commented May 3, 2015 at 15:03
  • $\begingroup$ @AlyssaWallace May you kindly describe your confusion again? By definition, you have an open set $U$. And open interval is a special case of open set. $\endgroup$
    – John
    Commented May 3, 2015 at 15:06
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    $\begingroup$ I guess I worry that if it is an open interval than the two "end points" aren't included. I am not even sure if that logically makes sense.. $\endgroup$
    – user219081
    Commented May 3, 2015 at 15:10
  • $\begingroup$ @AlyssaWallace Open interval is an open set for standard topology on $\mathbb{R}$. It may not be open for other topology. Your statement holds for general topological space, so I think you don't need to worry about which topology you are referring. Just prove it based on definition. $\endgroup$
    – John
    Commented May 3, 2015 at 15:13
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Note $x\in Int(A)$ and $Int(A)$ is open, because $Int(A)$ is the union of all open subsets of $A$ and the union of open sets is an open set. So, we can find an open subset $G\in \mathfrak T$ such that $x\in G\subseteq A$. Now consider that $A\subseteq B$, so $x\in G\subseteq B$. And that means $x\in Int(B)$.

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