Let $(X, \mathfrak T)$ be topological space and suppose that $A$ is a subset of $X$. Then $Bd(A) \subseteq A'$.

My definition of boundary: 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 $X−A$.

My definition of limit point: Let $(X, \mathfrak T)$ be a topological space with $ A \subseteq X$. A point $x \in X$ is said to be a limit point of $A$ provided that every open set containing $x$ contains a point $A$ different from $x$.

I am in an introduction to proofs writing class. I have to decide if this is true or false and then prove or provide a counter example. I think this is true and so I have started a proof. Our proofs on topology always deal with sets and elements in sets so I am trying to mirror that here.

I would like to start with the following: Let $x \in Bd(A)$ then by the definition $x \in A$ and $x \in X-A$. Now would it be correct to continue with a proof or a do I need a proof by contradiction?

  • $\begingroup$ I guess in line 5 you meant "... be a limit point of $A$ provided that every open set containing $x$ contains a point of $A$...". $$$$ Also: $Bd(A)=$ boundary of $A$? $A'=$ closure of $A$? $\endgroup$ – user228113 Apr 29 '15 at 2:12
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    $\begingroup$ It is not true that $x \in A$ and $x \in X - A$ (this is impossible). You would need to show that $x \in Bd(A)$ imples that $x$ is a limit point of $A$. Is this true? Why don't you try some examples, even in $\mathbb{R}$? $\endgroup$ – Pedro M. Apr 29 '15 at 2:14
  • $\begingroup$ You are correct that was a really dumb comment by me last night. I am still pondering this morning. $\endgroup$ – user219081 Apr 29 '15 at 15:43
  • $\begingroup$ I am considering the following example: Example. Consider the Euclidean space $(\mathbb R,\mathfrak U)$ and let $A = (0,1) \cup \{2\}$. Then $A′=[0,1]$, and $B(A)=\{0,1,2\}$. So does that make this false? $\endgroup$ – user219081 Apr 29 '15 at 16:23
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    $\begingroup$ Not every point in the boundary of a set $A$ is necessarily a limitpoint of $A$, and your comment provides indeed a valid counterexample of that. Another counterexample is $A=\{0\}$ where $A$ has no limitpoints (empty) and the boundary of $A$ coincides with $A$ (not empty). $\endgroup$ – drhab Apr 30 '15 at 10:24

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