1

Let $(a_{n})$ a sequence of positive numbers.It is given that $\inf A=0$,where $A=\{a_{n}:n \in N\}$.How can I show that that the sequence $(a_{n})$ has a decreasing subsequence that converges to $0$?

Supriyo
  • 6,119
  • 7
  • 32
  • 60
evinda
  • 7,823
  • 3
    Use the definition of $\inf$: For every $\epsilon>0$, there is an $x\in A$ with $x<\epsilon$. Proceed inductively. – David Mitra Feb 09 '14 at 10:50
  • And beware of the order. If you choose, for example an $a_{n_k}<\frac 1k$ which exists by the definition of $\inf$, then be sure to choose $a_{n_{k+1}}<\frac 1{k+1}$ but also that $n_{k+1}>n_k$. – dafinguzman Feb 09 '14 at 12:28

1 Answers1

0

Hints:

For each $\;n\in\Bbb N\;$ there exists $\;x_n\in\{x_m\}_{m\in\Bbb N}\;\;s.t.\;\;0\le x_n<0+\frac1n\;$

In the $\;2\le n-$th step take a peek at $\;\min \,\left\{x_{n-1},\frac1{n-1}\right\}\ldots\;$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287