Bolzano-Weierstrass Property

I used here the proof of the Bolzano-Weierstrass Property from the book "Elementary Real Analysis" by Thomson et al.. Note this theorem is different than the Bolzano-Weierstrass Theorem.

\begin{theorem} A set of real numbers $E$ is closed and bounded if and only if every sequence of points chosen from the set has a subsequence that converges to a point that belongs to $E$. \end{theorem}

\begin{corollary} A set of real numbers $E$ is closed and bounded if and only if every sequence of $E$ has a point of accumulation that belongs to $E$. \end{corollary}

\begin{proof} Suppose that $\E$ is both closed and bounded and let $\{x_n\}$ be a sequence of points chosen from $E$. Since $E$ is bounded this sequence $\{x_n\}$ must be bounded too. We apply the Bolzano-Weierstrass theorem for sequences to obtain a subsequence $\{x_{n_k}\}$ that is both monotonic and convergent. If $x_{n_k}\rightarrow z$ then there is $K$ such that $|x_{n_k}-z|<\epsilon$ for all $k\geq K$ and any positive $\epsilon$ . Since all the points of this subsequence belong to $E$ the neighborhoods of $z$ contain infinite points belong to $E$. So, by definition $z$ is a point of accumulation of $E$. Now we see that $ z\in E$ since $\E$ is closed.

So the Bolzano-Weierstrass Property is really that every sequence of a set of real numbers has a subsequence that converges to a point, or has a point of accumulation, that belongs to this set.

In the opposite direction we suppose that there is a set E which has the Bolzano-Weierstrass property but we don't know in advance if $E$ is closed and bounded. Then E cannot be unbounded. For example, if $E$ is unbounded then there is a sequence of points $\{x_n\}$ of $E$ with $x_n\rightarrow \infty$ or $x_n\rightarrow -\infty$ and no subsequence of that sequence converges, which contradicts the assumption. Also, $E$ must be closed. If $E$ is not closed, there is a point of accumulation $z$ such that $z\ni E$. This means that there is a sequence of points $\{x_n\}$ in $E$ converging to $z$. But any subsequence of $\{x_n\}$ would also converge to $z$ and $z\ni E$, which contradicts the Bolzano-Weierstrass Property assumed for $E$. \end{proof}

References: Bolzano-Weierstrass theorem

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