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proof

The above link is the proof of $\lim \inf x_n + \lim \inf y_n \le \lim \inf (x_n+y_n)$ when $x_n$ and $y_n$ are bounded.

Could you tell me why $XY_n \subset X_n+Y_n$??

Also, my textbook gives me a hint that 'Find a subsequence $\{x_{n_i}+y_{n_i}\}$ that converges. Then find a subsequence $\{x_{n_{m_i}}\}$ of $\{x_{n_i}\}$ that converges. Then apply what you know about limits.'. I really have no idea how to use this hint. Please tell me if you know how to use this hint.

Thank you in advance.

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  • $\begingroup$ Are you using Basic Analysis by Jeril Lebl? $\endgroup$ Nov 3, 2018 at 18:02

1 Answer 1

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For the first question: For $k\geq n$, $x_{k}+y_{k}\in X_{n}+Y_{n}$ because each $x_{k}\in X_{n}=\{x_{l}: l\geq n\}$ and $y_{k}\in Y_{n}=\{y_{l}: l\geq n\}$. So we conclude that $XY_{n}\subseteq X_{n}+Y_{n}$.

For the second question: Let $\liminf(x_{n}+y_{n})=\lim_{k}(x_{n_{k}}+y_{n_{k}})$. Since $(x_{n_{k}})$ is bounded, then there is a further subsequence $(x_{n_{k_{i}}})$ of $(x_{n_{k}})$ such that $\lim_{i}x_{n_{k_{i}}}$ exists. Since $\liminf x_{n}$ is the smallest limit point of its convergent subsequences, so $\liminf x_{n}\leq\lim_{i}x_{n_{k_{i}}}$. Now the corresponding subsequence $(y_{n_{k_{i}}})$ of $(y_{n_{k}})$ need no converge, but there is a further subsequence $(y_{n_{k_{i_{l}}}})$ of $(y_{n_{k_{i}}})$ such that $\lim_{l}y_{n_{k_{i_{l}}}}$ exists. Once again we have $\liminf y_{n}\leq\lim_{l}y_{n_{k_{i_{l}}}}$. The corresponding subsequence $(x_{n_{k_{i_{l}}}}+y_{n_{k_{i_{l}}}})$ of $(x_{n_{k}}+y_{n_{k}})$ is convergent and $\lim_{l}(x_{n_{k_{i_{l}}}}+y_{n_{k_{i_{l}}}})=\lim_{k}(x_{n_{k}}+y_{n_{k}})$. But $\lim_{l}(x_{n_{k_{i_{l}}}}+y_{n_{k_{i_{l}}}})=\lim_{l}x_{n_{k_{i_{l}}}}+\lim_{l}y_{n_{k_{i_{l}}}}=\lim_{i}x_{n_{k_{i}}}+\lim_{l}y_{n_{k_{i_{l}}}}$, the result follows.

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  • $\begingroup$ For the first question, why $XY_n = X_n+Y_n$does not hold?? $\endgroup$
    – shk910
    Mar 25, 2018 at 8:47
  • $\begingroup$ $X_{n}+Y_{n}$ may contain an element of the form $x_{n}+y_{n+1}$ which is not contained in $XY_{n}$. $\endgroup$
    – user284331
    Mar 25, 2018 at 8:51
  • $\begingroup$ Could you tell me why $y_{n_{k_i}}$ need no converge?? $\endgroup$
    – shk910
    Mar 25, 2018 at 8:59
  • $\begingroup$ And if we follow the last line of the proof, isn't it concluded as $\lim \inf x_n + \lim \inf y_n = \lim \inf (x_n+y_n)$? $\endgroup$
    – shk910
    Mar 25, 2018 at 9:02
  • $\begingroup$ No, because we have $\lim_{l}...\geq\liminf y_{n}$ for example. $\endgroup$
    – user284331
    Mar 25, 2018 at 16:03

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