2
$\begingroup$

Let $A = \{ \frac{1}{n} : n \in \Bbb{N} \}$, show that $\overline{A} = A \cup \{0\}.$

We have $A \subseteq \overline{A}$ by the definition of closures. To show that $\{0\} \subset \overline{A}$ we need to show that for every open set $U$ containing $0, U \cap A \not= \emptyset$. The elements of $A$ converge to $0$, since $\forall \epsilon > 0 \ \exists N \in \Bbb{N}, \ s.t \ \forall n > N \ \frac{1}{n} \in B_{\epsilon}(0).$ So that any open set $U$ containing $0$, we have an $\epsilon$ s.t $0 \in B_{\epsilon}(0) \subseteq U $. And since $B_{\epsilon}(0)$ belongs to $A$, we have that $A \cap U \not= \emptyset \ \forall \ U.$ This establishes that $A \cup \{0\} \subseteq \overline{A}$.

I'm not too sure how to show this direction $\overline{A} \subseteq A \cup \{0\}$ .

$\endgroup$
1
$\begingroup$

Here is another way to do it which I find easier:

Since $\overline{A}$ is defined as the set $A$ and all its limit points, by definition $\overline{A}=A\cup L$, where $L$ is the set of limit points of $A$.

Now you only need to show that $L=\{0\}$, i.e. that $A$ only has only $0$ as its limit point. The result then follows immediately.

$\endgroup$
1
$\begingroup$

Let $ a_n $ be a convergent sequence with its terms in $ \bar{A} $. If it has finitely many distinct elements, then it has to become constant after a certain index, and therefore its limit is in $ \bar{A} $. Otherwise, the sequence $ a_k $ contains arbitrarily small elements of $ A $, so there is a subsequence $ b_k $ converging to $ 0 $. Then, we have $ \lim_{n\to \infty} a_n = \lim_{n\to \infty} b_n = 0 $ as a sequence and its subsequences cannot converge to different limits. As every sequence with its terms in $ \bar{A} $ has its limit in $ \bar{A} $, the result follows.

$\endgroup$
1
$\begingroup$

It suffices to show that $\overline{A}-A=\{0\}$.

To show that $\overline A-A \subseteq \{0\}$ Suppose there exists some $a \in \overline{A}-A$ so that $a \neq 0$.

Suppose $a<0$. Then $\mathcal{B}(a,\frac{a}{2})$ shows a contradiction.

Supposes $a\geq 1$, and you will find a contradiction for the same reason.

Finally, suppose that $a \in (0,1)$. Then there exists $n \in \mathbb{N}$ so that $\frac{1}{n+1}<a<\frac{1}{n}$. Well, define an open ball $\frac{1}{2n^2+2n}$ around $a$ to show a contradiction.

There are slicker ways to reach this conclusion, but this is the most "Straightforward." (( for example, $\{\frac{1}{n}\}$ is a convergent sequence, so you can say something about any of its convergent subsequences )

$\endgroup$
  • $\begingroup$ Hence the assumption that $a \in \overline{A}-A$. $\endgroup$ – Andres Mejia May 28 '16 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.