Is the Limit of a Convergent Sequence in [0,1]?

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SUMMARY

The discussion centers on proving that the limit of a convergent sequence {an} contained within the interval [0,1] also lies within the same interval. The key argument involves the definition of limits, specifically that for every ε > 0, there exists a natural number N such that for all n > N, the absolute value |an - L| < ε holds true. The proof explores the implications of assuming L is outside [0,1], leading to contradictions when setting ε based on the values of L.

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  • Understanding of convergent sequences in real analysis
  • Familiarity with the epsilon-delta definition of limits
  • Knowledge of the properties of closed intervals in real numbers
  • Basic proof techniques in mathematical analysis
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  • Study the epsilon-delta definition of limits in detail
  • Explore properties of closed intervals in real analysis
  • Learn about convergence criteria for sequences
  • Review proof techniques, particularly proof by contradiction
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Homework Statement



suppose that {an} is a convergent sequence of points all in [0,1]. prove that lim an as n-->\infty is also in [0,1]

Homework Equations



for all\epsilon>0, \exists a natural number N such that for all natural numbers n, if n>N, then absolute value(an-L)<\epsilon

The Attempt at a Solution


i messed with the limit a lot, but the furthest i could get was that [0,1] was a subset of (-\epsilon,1+\epsilon), which contains L (the limit). can someone shed some light on this?
 
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Suppose L were NOT in [0, 1]. Then either L< 0 or L> 1.

If L< 0, let \epsilon= |L|/2.

If L> 1, let \epsilon= (L-1)/2.
 

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