$real analysis$

Question

$real analysis$

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

Real Analysis

Burgoos

226

File #1 (PNG)

Report

Burgoos

0

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

0

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

0

and hence has a convergent ...
Burgoos

0

thank you philip
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

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Burgoos

0

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

0

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

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

(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

4.6K

$real analysis$

Answer

Related Questions

Search