$real analysis$

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$real analysis$

check picture please, also its an hour and 30 mins not 2 hours

Real Analysis

Burgoos

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Burgoos

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philip, could you check my other questions if you its worth your time? one of them is very straightforward
Burgoos

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and hence ha a conveging subsequence - could you correct this?
Erdos

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and hence has a convergent ...
Burgoos

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thank you philip
Erdos

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Generally, the bounties should be higher for questions with short deadlines to give uses enough incentive to drop what they are doing and work on you problems.

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Burgoos

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philip, could you check my other questions if you its worth your time? one of them is very straightforward
Burgoos

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and hence ha a conveging subsequence - could you correct this?
Erdos

0

Generally, the bounties should be higher for questions with short deadlines to give uses enough incentive to drop what they are doing and work on you problems.

Answer 1

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(a) True. Let $X=f^{-1}(\{\ y_n: n \in \mathbb{N}\})$. Then $X \subset [a,b]$ and hence ha a conveging subsequence $\{x_n\}$. Since $f$ is continuous, we have
\[f(\lim_{n \rightarrow \infty } x_n)=\lim_{n \rightarrow \infty}f(x_n)= \lim_{n \rightarrow \infty}y_n.\]
(b) False. Let $U_n= (-\frac{1}{n}, \frac{1}{n})$. Then
\[\bigcap_{n=1}^{\infty}U_n=\{0\}\]
which is not open.
(c) True. Let $n\in \mathbb{N}$. For $\epsilon=\frac{1}{n}$ there exists $x_n$ such that
\[|x_n-a|<\frac{1}{n} \text{and} \ \ f(x_n)=\pi.\]
Hence $x_n$ converges to $x$, and since $f$ is continuous we have
\[f(a)=\lim_{n\rightarrow \infty} f(x_n)=\pi.\]

Erdos

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$real analysis$

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