Is $A = { f in C[0,1] mid f < alpha }$ an open subset in $C[0,1]$? The 2019 Stack Overflow...
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Is $A = { f in C[0,1] mid f
The 2019 Stack Overflow Developer Survey Results Are Inconvergence of $alpha$-Hölder-continuous functionsThe set ${|f|_alpha leq 1 }$ has compact closure in $C([0,1])$Proving that this set is open in a metric space.Show $f(x)=sqrt{x}$ is continuous on $[0,1]$Show that $[0,1)$ has no maximum, i.e. $not exists max[0,1)$Use the non-increasing and non-decreasing theorem to show that ${S_n= frac{alpha+ n}{beta + n}}$ converges.Find the limit , Show that ${F_n}$ converges uniformly to $F$ on closed subsets of $S$, but not on $S$. $F_n(x) = x^n sin nx, S=(-1,1)$Prove that $f:[0,1] to [0,1]$ has a fixed pointLet $0< alpha < beta leq 1$. Prove $Lip_{beta}[a,b] subset Lip_{alpha}[a,b]$.Proving that the interior of a metric space is open.
$begingroup$
Let $$epsilon = frac{1}{2} left| maxlimits_{x in [0,1]} f(x) - alpharight|$$
With $$mathrm{d}(f,g) = suplimits_{x in [0,1]} |f(x) - g(x)|$$,
and $f in A$. We only need to prove that every $h in B(f,epsilon)$ is an interior point. Pick one $ h in B(f, epsilon)$ and let’s make
$$
mathrm d(h,f) = d(h) < epsilon, \
epsilon_{h} = frac{epsilon - d(h)}{2},
$$ and let $i in B(h,epsilon_{h}).$
We have to show $mathrm{d}(i,f) < epsilon$:
$$d(i,f) leq d(i,h) + d(h,f) < epsilon_{h} + d(h) = frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
Then $$B(h,epsilon_{h}) subset B(f,epsilon) subset A rightarrow h in A rightarrow B(f,epsilon) subset A$$
Is this proof ok? How it changes if we talk about $C(0,1)$ instead $C[0,1]$?
How it changes if we prefer $L^1$ metric?
real-analysis
edited Mar 21 at 22:23
Brian
1,508416
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
$endgroup$
add a comment |
$begingroup$
Let $$epsilon = frac{1}{2} left| maxlimits_{x in [0,1]} f(x) - alpharight|$$
With $$mathrm{d}(f,g) = suplimits_{x in [0,1]} |f(x) - g(x)|$$,
and $f in A$. We only need to prove that every $h in B(f,epsilon)$ is an interior point. Pick one $ h in B(f, epsilon)$ and let’s make
$$
mathrm d(h,f) = d(h) < epsilon, \
epsilon_{h} = frac{epsilon - d(h)}{2},
$$ and let $i in B(h,epsilon_{h}).$
We have to show $mathrm{d}(i,f) < epsilon$:
$$d(i,f) leq d(i,h) + d(h,f) < epsilon_{h} + d(h) = frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
Then $$B(h,epsilon_{h}) subset B(f,epsilon) subset A rightarrow h in A rightarrow B(f,epsilon) subset A$$
Is this proof ok? How it changes if we talk about $C(0,1)$ instead $C[0,1]$?
How it changes if we prefer $L^1$ metric?
real-analysis
edited Mar 21 at 22:23
Brian
1,508416
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
$endgroup$
3
$begingroup$
In any metric space with the metric topology, it will always be true that ${x~|~||x|| lt alpha}$ is open. Your proof shows that.
$endgroup$
– Robert Shore
Mar 21 at 21:17
add a comment |
$begingroup$
Let $$epsilon = frac{1}{2} left| maxlimits_{x in [0,1]} f(x) - alpharight|$$
With $$mathrm{d}(f,g) = suplimits_{x in [0,1]} |f(x) - g(x)|$$,
and $f in A$. We only need to prove that every $h in B(f,epsilon)$ is an interior point. Pick one $ h in B(f, epsilon)$ and let’s make
$$
mathrm d(h,f) = d(h) < epsilon, \
epsilon_{h} = frac{epsilon - d(h)}{2},
$$ and let $i in B(h,epsilon_{h}).$
We have to show $mathrm{d}(i,f) < epsilon$:
$$d(i,f) leq d(i,h) + d(h,f) < epsilon_{h} + d(h) = frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
Then $$B(h,epsilon_{h}) subset B(f,epsilon) subset A rightarrow h in A rightarrow B(f,epsilon) subset A$$
Is this proof ok? How it changes if we talk about $C(0,1)$ instead $C[0,1]$?
How it changes if we prefer $L^1$ metric?
real-analysis
edited Mar 21 at 22:23
Brian
1,508416
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
$endgroup$
Let $$epsilon = frac{1}{2} left| maxlimits_{x in [0,1]} f(x) - alpharight|$$
With $$mathrm{d}(f,g) = suplimits_{x in [0,1]} |f(x) - g(x)|$$,
and $f in A$. We only need to prove that every $h in B(f,epsilon)$ is an interior point. Pick one $ h in B(f, epsilon)$ and let’s make
$$
mathrm d(h,f) = d(h) < epsilon, \
epsilon_{h} = frac{epsilon - d(h)}{2},
$$ and let $i in B(h,epsilon_{h}).$
We have to show $mathrm{d}(i,f) < epsilon$:
$$d(i,f) leq d(i,h) + d(h,f) < epsilon_{h} + d(h) = frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$
Then $$B(h,epsilon_{h}) subset B(f,epsilon) subset A rightarrow h in A rightarrow B(f,epsilon) subset A$$
Is this proof ok? How it changes if we talk about $C(0,1)$ instead $C[0,1]$?
How it changes if we prefer $L^1$ metric?
real-analysis
real-analysis
edited Mar 21 at 22:23
Brian
1,508416
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
edited Mar 21 at 22:23
Brian
1,508416
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
edited Mar 21 at 22:23
Brian
1,508416
edited Mar 21 at 22:23
Brian
1,508416
edited Mar 21 at 22:23
Brian
1,508416
1,508416
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
asked Mar 21 at 20:53
Pablo Valentin Cortes CastilloPablo Valentin Cortes Castillo
477
477
3
$begingroup$
In any metric space with the metric topology, it will always be true that ${x~|~||x|| lt alpha}$ is open. Your proof shows that.
$endgroup$
– Robert Shore
Mar 21 at 21:17
add a comment |
3
$begingroup$
In any metric space with the metric topology, it will always be true that ${x~|~||x|| lt alpha}$ is open. Your proof shows that.
$endgroup$
– Robert Shore
Mar 21 at 21:17
3
3
$begingroup$
In any metric space with the metric topology, it will always be true that ${x~|~||x|| lt alpha}$ is open. Your proof shows that.
$endgroup$
– Robert Shore
Mar 21 at 21:17
$begingroup$
In any metric space with the metric topology, it will always be true that ${x~|~||x|| lt alpha}$ is open. Your proof shows that.
$endgroup$
– Robert Shore
Mar 21 at 21:17
add a comment |
0
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$begingroup$
In any metric space with the metric topology, it will always be true that ${x~|~||x|| lt alpha}$ is open. Your proof shows that.
$endgroup$
– Robert Shore
Mar 21 at 21:17