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I'm taking a first course in real analysis, and we're using Rudin's Principles of Mathematical Analysis as our main (only) book.

In chapter two, Rudin discusses basic topology from the point of view of metric spaces, and one of the concepts defined in the book is connectedness, with he following definition being given:

Let $(X,d)$ be a metric space. Two sets $A\subset X$ , $B\subset X$ are said to be separated iff $\bar{A} \cap{B} = \varnothing$ and $\bar{B} \cap{A} = \varnothing$, where $\bar{A}$ denotes the closure of A.

A set $E \subset X$ is said to be connected iff E is NOT the union of two non-empty separated sets.

Now, reading up elsewhere I've come across the following theorem:

Let $(X,d)$ be a metric space, and let $E \subset X$. The following are equivalent:

(1) E is the union of two non-empty separated sets.

(2) There exist $A,B \subset X$, $A\neq\varnothing$, $B\neq\varnothing$, $A \cap{B}=\varnothing$ A, B open relative to E.

(3) There exist $A,B \subset X$, $A\neq\varnothing$, $B\neq\varnothing$, $A \cap{B}=\varnothing$ A, B closed relative to E.

Note: A set $A, A \subset Y \subset X$ is said to be open relative to $Y$ iff:

$$(\forall p\in A)(\exists \varepsilon>0)((B_{\varepsilon}(p)\cap Y) \subset A)$$

Now, I've been able to prove $(1)\Rightarrow(2)$ and $(2)\Rightarrow(3)$ but I'm unable to show that $(3)\Rightarrow(1)$ (which would complete the proof).

Does anyone have any suggestions?

Thanks!

PS: Rudin defines the closure of a set in the following way.

Let $(X,d)$ be a metric space, and let $E \subset X$.

$p \in X$ is said to be a limit point of E iff:

$$(\forall \varepsilon>0)((B_{\varepsilon}(p)\cap E) \setminus \left \{ p \right \} \neq \varnothing)$$

Let $E'$ be the set of all limit points of E, then define $\bar{E} = E \cup E´$. It then follows that $E$ is closed if and only if $\bar{E}=E$, and that \bar{E} is the smallest closed set that contains $E$.

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You may want to show that, if $A\subset E$ and we denote the closure of $A$ in $E$ by $\bar{A}^E$, then $\bar{A}^E=\bar{A}\cap E$ where $\bar{A}$ denotes the closure of $A$ in $X$. So, if $A$ is closed in $E$, $A=\bar{A}\cap E$. It is also useful to note that $B=B\cap E$ since $B\subset E$.

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  • $\begingroup$ Great, I think that does it, I hadn't noticed that property regarding the closure of A in E. Thank you very much! $\endgroup$ – Reveillark Sep 20 '14 at 3:59

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