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

If $A\subseteq B$ then $A' \subseteq B'$ ($A'$ is the set of limit points)

Let $( X, \mathfrak T_U)$ be the topological space. Let $A = [0,1) \cup (1,2)$ Let $B = [0,1) \cup (1,3)$

Then $A\subseteq B$ but $Bd(A)= \{0,1,2\}$ and $Bd(B) = \{0,1,3\}$ therefore $Bd(A) \not \subseteq Bd(B)$ so this is a false conjecture.

I am a little more uncomfortable with the limit points. I believe $A'= [0,1] \cup [2, \infty)$ and $B'= [0,1] \cup [3, \infty)$ and therefore again this would be a false conjecture.

My definition of limit point is that every open set containing $x$ contains a point of $A$ different from $x$.

My definition of boundary is: Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of $A $if every open set containing $x$ intersects both $A$ and $X−A$

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  • $\begingroup$ What is $A'$? ${}{}$ $\endgroup$ – copper.hat May 12 '15 at 1:56
  • $\begingroup$ Do primes denote boundary or something? Maybe closure? $\endgroup$ – Alfred Yerger May 12 '15 at 1:56
  • $\begingroup$ Sorry $A'$ is the set of limit points $\endgroup$ – user219081 May 12 '15 at 2:02
  • $\begingroup$ The boundary of $A$ is $\{0,1,2\}$, the boundary of $B$ is $\{0,1,3\}$. The limit points of $A$ are $[0,2]$. The limit points of $B$ are $[0,3]$. $\endgroup$ – copper.hat May 12 '15 at 2:09
  • $\begingroup$ @cooper.hat thank you for fixing that mistake. Will this always be true of the limit points? $\endgroup$ – user219081 May 12 '15 at 2:20
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Remember the definition for $D'$, $D\subseteq X$:

$$D'=\{x\in X/\ \forall V\in \mathfrak T(x\in V\implies \exists d\in D(d\neq x \wedge d\in V))\}$$

(I'm sorry for the symbolic notation), let's prove that $A'\subseteq B'$ if $A\subseteq B$.

Let $x\in A'$ and take any $V\in \mathfrak T$ such that $x\in V$, since $x\in A'$ there exists $a\in A$ such that $a\neq x$ and $a\in V$, in particular, if we let $b:=a$, we see that $b\in B$ (since $A\subseteq B$), $b\neq x$ and $b\in V$. This shows that

$$\forall V\in \mathfrak T(x\in V\implies \exists b\in D(b\neq x \wedge b\in V))$$

so $x\in B'$. Since $x\in A'$ is arbitrary, we have that $A'\subseteq B'$.

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If $a_n \to x$, $a_n \in A$ and $a_n \neq x$, then we also have $a_n \in B$.

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Let $x$ be a limit point of $A$. Then for any open set $U$ with $x \in U$ we know $$\left[U\setminus \{x\}\right]\cap A \neq \emptyset$$ Since $A\subset B$ then $$\left[U\setminus \{x\}\right]\cap A \subset \left[U\setminus \{x\}\right]\cap B$$ and hence $$\left[U\setminus \{x\}\right]\cap B \neq \emptyset$$ so every limit point of $A$ is a limit point of $B$.

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  • $\begingroup$ Can you explain the $[U \setminus \{x\}] \cap A =\emptyset$ $\endgroup$ – user219081 May 13 '15 at 19:30
  • $\begingroup$ @AlyssaWallace No problem, that is just the/a definition of a limit point. Given a set $C$, $x$ is a limit point of $C$ if and only any open set $U$ that contains $x$ is such that $[U\setminus \{x\}]\cap C$ is nonempty. Are you using a different definition of limit point? $\endgroup$ – graydad May 13 '15 at 19:35
  • $\begingroup$ My definition of limit point is that every open set containing $x$ contains a point of $A$ different from $x$. $\endgroup$ – user219081 May 13 '15 at 19:37
  • $\begingroup$ @AlyssaWallace yep, that is the same definition. Since $[U\setminus \{x\}]\cap C$ is nonempty and $x \notin [U\setminus \{x\}]\cap C$ then there must be some other point in $C$ besides $x$ that is also contained in $U$ $\endgroup$ – graydad May 13 '15 at 19:40

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