- #1
Ahmad Kishki
- 159
- 13
Homework Statement
Show that the set of limit points of the set ##A \subseteq \mathbb{R}## given by ##L## is a closed set.2. Relevant theorems
Definition: A closed set F (subset of R) is such that it contains its limit points.
Definition: A limit point x of a set A (subset of R) is such that the intersection Of every epsilon neighborhood with A excluding x is not empty.
Theorem: A number x is a limit point if and only if some cauchy sequence in A is convergent to x while each term in that cauchy sequence is not equal to x.
Theorem: A set F is closed if and only if every cauchy sequence in F has a limit that is also an element of F.
The Attempt at a Solution
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Proof: Since ##L## is the set of limit points ##A \subseteq \mathbb{R}##, then if ##(a_n)## is a cauchy sequence in ##A## then ##\lim a_n \in L##. Suppose ##x## is a limit point of ##L##, then we can form a cauchy sequence satisfying ##l_n \neq x## for all ##n##, and ##\lim l_n = x##. However, each term in the sequence ##(l_n)## is a limit reached by some cauchy sequence in ##A##.
I will try to construct a cauchy sequence in A that converges to x hence showin that ##x \in L## and ##L## is a closed set.
We start by choosing ##(a_n)## such that;
$$|a_n - l_n| < \frac{1}{n}$$
For this definition of ##(a_n)##, given any ##\epsilon > 0##, we choose ##\frac{1}{N_1} < \frac{\epsilon}{2}## such that for all ##n \geq N_1##;
$$|a_n - l_n| < \frac{\epsilon}{2}$$
Since ##(l_n)## converges to ##x##, then, for all ##n \geq N_2##;
$$|l_n - x| < \frac{\epsilon}{2}$$
Choosing ##N = \max{(N_1,N_2)}##;
$$|a_n - x| = |(a_n - l_n) - (l_n - x)| \leq |a_n - l_n| + |l_n - x| < \epsilon$$
Hence, this contruction of ##(a_n)## converges to ##x##.
Q.E.D.
Is this correct?
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