If $(k_n)$ is a sequence in $K subseteq mathbb{R}$ such that $k_n to k_0 notin K$, then there exists an...
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4
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Here's the question:
Let $Ksubseteq mathbb{R}$ and suppose that there exists a sequence
$(k_n)$ in $K$ that converges to a number $k_0 notin K$. Show that
there is an unbounded continuous function on $K$.
Here's the function I choose: $fcolon K to mathbb{R}$ given by $f(x)=frac{1}{k_0 - x}$. Clearly, f is continuous and it remains to show $f$ is unbounded. Suppose that $f$ is bounded i.e. $|f(x)|< M$ for all $x in K$ (also for some $M>0$). This means that that $|f(k_n)| < M $ for all $n in mathbb{N}$. Hence, $lim |f(k_n)| le M$ but then this is a contradiction since $lim |f(k_n)| = +infty$.
Is this proof okay? And I've chosen my codomain to be the $mathbb{R}$ and is that fine?
real-analysis sequences-and-series proof-verification continuity
edited Nov 17 at 5:59
Brahadeesh
5,78441957
asked Jul 1 at 9:40
Ashish K
775513
add a comment |
up vote
4
down vote
favorite
Here's the question:
Let $Ksubseteq mathbb{R}$ and suppose that there exists a sequence
$(k_n)$ in $K$ that converges to a number $k_0 notin K$. Show that
there is an unbounded continuous function on $K$.
Here's the function I choose: $fcolon K to mathbb{R}$ given by $f(x)=frac{1}{k_0 - x}$. Clearly, f is continuous and it remains to show $f$ is unbounded. Suppose that $f$ is bounded i.e. $|f(x)|< M$ for all $x in K$ (also for some $M>0$). This means that that $|f(k_n)| < M $ for all $n in mathbb{N}$. Hence, $lim |f(k_n)| le M$ but then this is a contradiction since $lim |f(k_n)| = +infty$.
Is this proof okay? And I've chosen my codomain to be the $mathbb{R}$ and is that fine?
real-analysis sequences-and-series proof-verification continuity
edited Nov 17 at 5:59
Brahadeesh
5,78441957
asked Jul 1 at 9:40
Ashish K
775513
3
yes, that looks fine.
– Thomas
Jul 1 at 9:44
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Here's the question:
Let $Ksubseteq mathbb{R}$ and suppose that there exists a sequence
$(k_n)$ in $K$ that converges to a number $k_0 notin K$. Show that
there is an unbounded continuous function on $K$.
Here's the function I choose: $fcolon K to mathbb{R}$ given by $f(x)=frac{1}{k_0 - x}$. Clearly, f is continuous and it remains to show $f$ is unbounded. Suppose that $f$ is bounded i.e. $|f(x)|< M$ for all $x in K$ (also for some $M>0$). This means that that $|f(k_n)| < M $ for all $n in mathbb{N}$. Hence, $lim |f(k_n)| le M$ but then this is a contradiction since $lim |f(k_n)| = +infty$.
Is this proof okay? And I've chosen my codomain to be the $mathbb{R}$ and is that fine?
real-analysis sequences-and-series proof-verification continuity
edited Nov 17 at 5:59
Brahadeesh
5,78441957
asked Jul 1 at 9:40
Ashish K
775513
Here's the question:
Let $Ksubseteq mathbb{R}$ and suppose that there exists a sequence
$(k_n)$ in $K$ that converges to a number $k_0 notin K$. Show that
there is an unbounded continuous function on $K$.
Here's the function I choose: $fcolon K to mathbb{R}$ given by $f(x)=frac{1}{k_0 - x}$. Clearly, f is continuous and it remains to show $f$ is unbounded. Suppose that $f$ is bounded i.e. $|f(x)|< M$ for all $x in K$ (also for some $M>0$). This means that that $|f(k_n)| < M $ for all $n in mathbb{N}$. Hence, $lim |f(k_n)| le M$ but then this is a contradiction since $lim |f(k_n)| = +infty$.
Is this proof okay? And I've chosen my codomain to be the $mathbb{R}$ and is that fine?
real-analysis sequences-and-series proof-verification continuity
real-analysis sequences-and-series proof-verification continuity
edited Nov 17 at 5:59
Brahadeesh
5,78441957
asked Jul 1 at 9:40
Ashish K
775513
edited Nov 17 at 5:59
Brahadeesh
5,78441957
asked Jul 1 at 9:40
Ashish K
775513
edited Nov 17 at 5:59
Brahadeesh
5,78441957
edited Nov 17 at 5:59
Brahadeesh
5,78441957
edited Nov 17 at 5:59
Brahadeesh
5,78441957
5,78441957
asked Jul 1 at 9:40
Ashish K
775513
asked Jul 1 at 9:40
Ashish K
775513
asked Jul 1 at 9:40
Ashish K
775513
775513
3
yes, that looks fine.
– Thomas
Jul 1 at 9:44
add a comment |
3
yes, that looks fine.
– Thomas
Jul 1 at 9:44
3
3
yes, that looks fine.
– Thomas
Jul 1 at 9:44
yes, that looks fine.
– Thomas
Jul 1 at 9:44
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Yes, it is correct. But you don't need to do it by contradiction. Given $Minmathbb{R}^+$, there is some $ninmathbb N$ such that $|k_0-k_n|<frac1M$. And this means that $bigl|f(k_n)bigr|>M$. Since this occurs for every $Minmathbb{R}^+$, $f$ is unbounded.
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, it is correct. But you don't need to do it by contradiction. Given $Minmathbb{R}^+$, there is some $ninmathbb N$ such that $|k_0-k_n|<frac1M$. And this means that $bigl|f(k_n)bigr|>M$. Since this occurs for every $Minmathbb{R}^+$, $f$ is unbounded.
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
add a comment |
up vote
1
down vote
accepted
Yes, it is correct. But you don't need to do it by contradiction. Given $Minmathbb{R}^+$, there is some $ninmathbb N$ such that $|k_0-k_n|<frac1M$. And this means that $bigl|f(k_n)bigr|>M$. Since this occurs for every $Minmathbb{R}^+$, $f$ is unbounded.
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, it is correct. But you don't need to do it by contradiction. Given $Minmathbb{R}^+$, there is some $ninmathbb N$ such that $|k_0-k_n|<frac1M$. And this means that $bigl|f(k_n)bigr|>M$. Since this occurs for every $Minmathbb{R}^+$, $f$ is unbounded.
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
Yes, it is correct. But you don't need to do it by contradiction. Given $Minmathbb{R}^+$, there is some $ninmathbb N$ such that $|k_0-k_n|<frac1M$. And this means that $bigl|f(k_n)bigr|>M$. Since this occurs for every $Minmathbb{R}^+$, $f$ is unbounded.
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
answered Jul 1 at 9:55
José Carlos Santos
142k20112208
142k20112208
add a comment |
add a comment |
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3
yes, that looks fine.
– Thomas
Jul 1 at 9:44