Mixing
If $f(a_{n_k}) to a$ as $k to infty$, then there is some $x in mathbb{R}$ such that $f(x) = a$ and $a_{n_k}...
$begingroup$
I have two questions:
Let $(a_n)_{n in mathbb{N}}$ be a bounded sequence and let $f: mathbb{R} to mathbb{R}$ be a continuous function.
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ there is there some $y in mathbb{R}$ such that $f(y) = a$ and $y$ is an accumulation point for $(a_n)$?
And also:
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ and there is some $y in mathbb{R}$ such that $f(y) = a$, then $y$ is an accumulation point for $(a_n)$.
I have tried writing the definition of continuity with $varepsilon - delta$ and with sequences, but I did not manage to solve the above problems, because that definition gives no information on $(a_n)$.
I have also tried constructing a counter example for the first question (since I believe to be false), but I was not able to.
Thank you in advance for the help!
real-analysis sequences-and-series continuity
asked Jan 27 at 18:53
S.T.S.T.
82
$endgroup$
add a comment |
$begingroup$
I have two questions:
Let $(a_n)_{n in mathbb{N}}$ be a bounded sequence and let $f: mathbb{R} to mathbb{R}$ be a continuous function.
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ there is there some $y in mathbb{R}$ such that $f(y) = a$ and $y$ is an accumulation point for $(a_n)$?
And also:
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ and there is some $y in mathbb{R}$ such that $f(y) = a$, then $y$ is an accumulation point for $(a_n)$.
I have tried writing the definition of continuity with $varepsilon - delta$ and with sequences, but I did not manage to solve the above problems, because that definition gives no information on $(a_n)$.
I have also tried constructing a counter example for the first question (since I believe to be false), but I was not able to.
Thank you in advance for the help!
real-analysis sequences-and-series continuity
asked Jan 27 at 18:53
S.T.S.T.
82
$endgroup$
add a comment |
$begingroup$
I have two questions:
Let $(a_n)_{n in mathbb{N}}$ be a bounded sequence and let $f: mathbb{R} to mathbb{R}$ be a continuous function.
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ there is there some $y in mathbb{R}$ such that $f(y) = a$ and $y$ is an accumulation point for $(a_n)$?
And also:
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ and there is some $y in mathbb{R}$ such that $f(y) = a$, then $y$ is an accumulation point for $(a_n)$.
I have tried writing the definition of continuity with $varepsilon - delta$ and with sequences, but I did not manage to solve the above problems, because that definition gives no information on $(a_n)$.
I have also tried constructing a counter example for the first question (since I believe to be false), but I was not able to.
Thank you in advance for the help!
real-analysis sequences-and-series continuity
asked Jan 27 at 18:53
S.T.S.T.
82
$endgroup$
I have two questions:
Let $(a_n)_{n in mathbb{N}}$ be a bounded sequence and let $f: mathbb{R} to mathbb{R}$ be a continuous function.
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ there is there some $y in mathbb{R}$ such that $f(y) = a$ and $y$ is an accumulation point for $(a_n)$?
And also:
If $a$ is an limit point for $f(a_n)_{n in mathbb{N}},$ and there is some $y in mathbb{R}$ such that $f(y) = a$, then $y$ is an accumulation point for $(a_n)$.
I have tried writing the definition of continuity with $varepsilon - delta$ and with sequences, but I did not manage to solve the above problems, because that definition gives no information on $(a_n)$.
I have also tried constructing a counter example for the first question (since I believe to be false), but I was not able to.
Thank you in advance for the help!
real-analysis sequences-and-series continuity
real-analysis sequences-and-series continuity
asked Jan 27 at 18:53
S.T.S.T.
82
asked Jan 27 at 18:53
S.T.S.T.
82
asked Jan 27 at 18:53
S.T.S.T.
82
asked Jan 27 at 18:53
S.T.S.T.
82
asked Jan 27 at 18:53
S.T.S.T.
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
If $a$ is a limit point of ${f(a_n)}$, then there exists a subsequence $a_{k_n}$, such that
$$
f(a_{k_n})to a
$$
and since ${a_{n_k}}$ is bounded, then it possesses a convergent subsequence, say $a_{ell_n}to b$. Now we have that
$$
a_{ell_n}to bqquadtext{and}qquad f(a_{ell_n})to f(b)
$$
and also $f(a_{ell_n})to a$.
Thus $f(b)=a$.
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $a$ is a limit point of ${f(a_n)}$, then there exists a subsequence $a_{k_n}$, such that
$$
f(a_{k_n})to a
$$
and since ${a_{n_k}}$ is bounded, then it possesses a convergent subsequence, say $a_{ell_n}to b$. Now we have that
$$
a_{ell_n}to bqquadtext{and}qquad f(a_{ell_n})to f(b)
$$
and also $f(a_{ell_n})to a$.
Thus $f(b)=a$.
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
add a comment |
$begingroup$
If $a$ is a limit point of ${f(a_n)}$, then there exists a subsequence $a_{k_n}$, such that
$$
f(a_{k_n})to a
$$
and since ${a_{n_k}}$ is bounded, then it possesses a convergent subsequence, say $a_{ell_n}to b$. Now we have that
$$
a_{ell_n}to bqquadtext{and}qquad f(a_{ell_n})to f(b)
$$
and also $f(a_{ell_n})to a$.
Thus $f(b)=a$.
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
add a comment |
$begingroup$
If $a$ is a limit point of ${f(a_n)}$, then there exists a subsequence $a_{k_n}$, such that
$$
f(a_{k_n})to a
$$
and since ${a_{n_k}}$ is bounded, then it possesses a convergent subsequence, say $a_{ell_n}to b$. Now we have that
$$
a_{ell_n}to bqquadtext{and}qquad f(a_{ell_n})to f(b)
$$
and also $f(a_{ell_n})to a$.
Thus $f(b)=a$.
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
$endgroup$
If $a$ is a limit point of ${f(a_n)}$, then there exists a subsequence $a_{k_n}$, such that
$$
f(a_{k_n})to a
$$
and since ${a_{n_k}}$ is bounded, then it possesses a convergent subsequence, say $a_{ell_n}to b$. Now we have that
$$
a_{ell_n}to bqquadtext{and}qquad f(a_{ell_n})to f(b)
$$
and also $f(a_{ell_n})to a$.
Thus $f(b)=a$.
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
answered Jan 27 at 20:24
Yiorgos S. SmyrlisYiorgos S. Smyrlis
63.6k1385165
63.6k1385165
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
add a comment |
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
$begingroup$
Thank you for the answer!
$endgroup$
– S.T.
Jan 27 at 20:31
add a comment |
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