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Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ is a subset of $X$. If $Bd(A) = \emptyset$ then $A = \emptyset$ or $A = X$.

I am studying introduction to proofs and we have learned set theory and a basic introduction to topology in order to write proofs.

Here are my definitions for this problem:

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 $X−A$

I have an example which demonstrates one of the conditions: Let $X = \mathbb R$ and $\mathfrak T = \{U \subseteq \mathbb R: 1 \in U$ or $U = \emptyset \}$ Let $ A =\{3,4\}$ Then $Bd(A) = \{3,4\}$ based on this definition I think this is a true statement so I know need to prove it. All of my proofs for topology involve sets and elements of sets. I only have a basic, introduction to topology.

As far as a proof goes I was thinking about a proof by contraction? But I do not know if that is a good way to go about it.

Proof: Assume $Bd(A) \neq \emptyset$ and $A = \emptyset$ or $A =X$ then there exists an $x \in X$ and therefore $x \in A$ therefore $A \neq \emptyset$.

Am I on the right track? How do I show the second condition $X = A$?

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  • $\begingroup$ It's an interesting topology in your example. But one rather pathological example shouldn't convince you that the general result is true. It's not completely obvious $\endgroup$ Commented May 2, 2015 at 14:20
  • $\begingroup$ Assuming that $\operatorname{Bd}(A)\neq \emptyset$, is wrong, since this is part of the setting of the excercise (the same mistake would be assuming that $A$ is not a subset of $X$, for example). If you want a proof by contradiction, you must assume that given conditions do not imply $A = \emptyset$ or $A = X$, that is, you are (additionaly) assuming that $A\neq \emptyset$ and $A\neq X$. $\endgroup$
    – Antoine
    Commented May 2, 2015 at 14:23
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    $\begingroup$ Following up on my previous comment, the statement is actually not true - without another assumption like connectedness - so you see what I mean about jumping to conclusions from one example. $\endgroup$ Commented May 2, 2015 at 14:23
  • $\begingroup$ @GregoryGrant is there example that shows this is not true? $\endgroup$
    – user219081
    Commented May 2, 2015 at 15:03

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The topology space $(X,\mathcal{T})$ should be connected. For a subset $A\subset X$ we have $Bd(A)=Cl(A)-A^{0}$. So, if $Bd(A)=\emptyset\Rightarrow Cl(A)=A^{0}=A$ Then $A$ is both open and closed. Since $X$ is connected, we have $A=X$ or $A=\emptyset$.

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  • $\begingroup$ Right I was just thinking that, if the space had two components like $(0,1)\cup(2,3)$ then the boundary of $(0,1)$ is indeed empty. $\endgroup$ Commented May 2, 2015 at 14:22
  • $\begingroup$ What is the definition of $A^0$ used here? $\endgroup$ Commented May 2, 2015 at 14:24
  • $\begingroup$ $A^\circ$ is the interior of $A$, the union of all open sets contained in $A$. $\endgroup$ Commented May 2, 2015 at 14:24
  • $\begingroup$ @Uncountable the interior of A. $\endgroup$
    – Jay
    Commented May 2, 2015 at 14:24
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    $\begingroup$ @AlyssaWallace I posted an example above, if $X=(0,1)\cup(2,3)$ (open intervals in $\mathbb R$) and $A=(0,1)$ then the boundary of $A$ is the empty set. $\endgroup$ Commented May 2, 2015 at 15:04

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