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Let [itex](F_n)[/itex] be a decreasing sequence [i.e., F_1 contains F2 which contains F_3...etc] of closed bounded nonempty sets in [itex]R^k[/itex]. Then [itex]F = \cap^{\infty}_{m=1} F_n[/itex] is also closed, bounded and nonempty.

The proof (from our textbook) says:

Clearly F is closed and bounded. It is the nonemptiness that needs proving! For each n, select an element [itex](

Why does it suffice to show that?

If [itex]m \geq n_0[/itex], then [itex]n_m \geq n_0[/itex], so [itex]

But what if m<n_0?

Hence the sequence [itex]{

Thanks in advance

The proof (from our textbook) says:

Clearly F is closed and bounded. It is the nonemptiness that needs proving! For each n, select an element [itex](

**x**_n)[/itex] in [itex]F_n[/itex]. By the Bolzano-Weierstrass theorem 13.5, a subsequence [itex](**x**_n_m})^{\infty}_{m=1}[/itex] of [itex](**x**_n)[/itex] converges to some element [itex]**x**_0[/itex] in [itex]R^k[/itex]. To show [itex]**x**_0 \in F[/itex], it suffices to show [itex]**x**_0 \in F_n_0[/itex] with [itex]n_0[/itex] fixed.Why does it suffice to show that?

If [itex]m \geq n_0[/itex], then [itex]n_m \geq n_0[/itex], so [itex]

**x**_n_m \in F_n_m \subseteq F_n_0[/itex].But what if m<n_0?

Hence the sequence [itex]{

**x**_n_m}^{\infty}_{m=1}[/itex] consists of points in [itex]F_n_0[/itex] and converges to [itex]**x**_0[/itex]. Thus [itex]**x**_0[/itex] belongs to [itex]F_n_0[/itex] by (b) of proposition 13.9 (which says “The set E is closed if and only if it contains the limit ofevery convergent sequence of points in E.)Thanks in advance

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