Showing that a set is meager
$begingroup$
Consider $X =C([-1,1])$ with the usual norm $lVert f rVert_{infty} = sup_{tin [-1,1]} |f(t)|.$ Let
begin{align*}
A_+ &= { f in X: f(t) = f(-t) quad forall t in[-1,1] }\
A_- &= { f in X: f(t) = -f(-t) quad forall tin [-1,1] }.
end{align*}
Show that $A_+$ and $A_{-}$ are meager.
I'm uncertain on how to proceed with this.
real-analysis general-topology functional-analysis functions baire-category
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
$endgroup$
add a comment |
$begingroup$
Consider $X =C([-1,1])$ with the usual norm $lVert f rVert_{infty} = sup_{tin [-1,1]} |f(t)|.$ Let
begin{align*}
A_+ &= { f in X: f(t) = f(-t) quad forall t in[-1,1] }\
A_- &= { f in X: f(t) = -f(-t) quad forall tin [-1,1] }.
end{align*}
Show that $A_+$ and $A_{-}$ are meager.
I'm uncertain on how to proceed with this.
real-analysis general-topology functional-analysis functions baire-category
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
$endgroup$
3
$begingroup$
You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed?
$endgroup$
– Mathematician 42
Dec 9 '17 at 21:03
2
$begingroup$
As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior.
$endgroup$
– Martin Sleziak
Dec 10 '17 at 11:20
add a comment |
$begingroup$
Consider $X =C([-1,1])$ with the usual norm $lVert f rVert_{infty} = sup_{tin [-1,1]} |f(t)|.$ Let
begin{align*}
A_+ &= { f in X: f(t) = f(-t) quad forall t in[-1,1] }\
A_- &= { f in X: f(t) = -f(-t) quad forall tin [-1,1] }.
end{align*}
Show that $A_+$ and $A_{-}$ are meager.
I'm uncertain on how to proceed with this.
real-analysis general-topology functional-analysis functions baire-category
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
$endgroup$
Consider $X =C([-1,1])$ with the usual norm $lVert f rVert_{infty} = sup_{tin [-1,1]} |f(t)|.$ Let
begin{align*}
A_+ &= { f in X: f(t) = f(-t) quad forall t in[-1,1] }\
A_- &= { f in X: f(t) = -f(-t) quad forall tin [-1,1] }.
end{align*}
Show that $A_+$ and $A_{-}$ are meager.
I'm uncertain on how to proceed with this.
real-analysis general-topology functional-analysis functions baire-category
real-analysis general-topology functional-analysis functions baire-category
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
edited Dec 10 '17 at 11:17
Martin Sleziak
44.7k9117272
44.7k9117272
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
asked Dec 9 '17 at 20:52
KalypsoKalypso
1917
1917
3
$begingroup$
You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed?
$endgroup$
– Mathematician 42
Dec 9 '17 at 21:03
2
$begingroup$
As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior.
$endgroup$
– Martin Sleziak
Dec 10 '17 at 11:20
add a comment |
3
$begingroup$
You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed?
$endgroup$
– Mathematician 42
Dec 9 '17 at 21:03
2
$begingroup$
As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior.
$endgroup$
– Martin Sleziak
Dec 10 '17 at 11:20
3
3
$begingroup$
You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed?
$endgroup$
– Mathematician 42
Dec 9 '17 at 21:03
$begingroup$
You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed?
$endgroup$
– Mathematician 42
Dec 9 '17 at 21:03
2
2
$begingroup$
As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior.
$endgroup$
– Martin Sleziak
Dec 10 '17 at 11:20
$begingroup$
As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior.
$endgroup$
– Martin Sleziak
Dec 10 '17 at 11:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each of $A_{+}$ and $A_{-}$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.
First, we note what an $varepsilon$-ball around a point $f$ looks like. It is the set of all continuous functions that differ from $f$ by less than $varepsilon$ at any point in $[0,1]$. That is, the set of of all continuous functions that lie within a $varepsilon$ “tube” around $f$.
To see that the $A_{+}$ is closed, we show its complement $U$, is open. If $g in U$, then $exists T in [0,1]$ such that $g(-T) neq g(T)$. Define $delta = g(T)-g(-T)$. Now for any continuous function $h$ in the $frac{delta}{3}$ ball around $g$, it’s clear that $h(T) neq h(-T)$. In other words, the entire $frac{delta}{3}$ ball around $g$ lies within $U$. So $U$ is open and therefore, $A_{+}$ is closed.
To show its interior is empty, pick any $fin A_{+}$ and consider an $varepsilon$-ball $B$ around $f$. Since this is an $varepsilon$-tube around $ f$, it is easy to see that there are continuous functions $sin B$ such that for some $T, s(T) neq s(-T)$. In other words, $A_{+}$ does not contain any open ball. So it has an empty interior.
The arguments are analogous for $A_{-}$.
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
$endgroup$
1
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
add a comment |
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$begingroup$
For each of $A_{+}$ and $A_{-}$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.
First, we note what an $varepsilon$-ball around a point $f$ looks like. It is the set of all continuous functions that differ from $f$ by less than $varepsilon$ at any point in $[0,1]$. That is, the set of of all continuous functions that lie within a $varepsilon$ “tube” around $f$.
To see that the $A_{+}$ is closed, we show its complement $U$, is open. If $g in U$, then $exists T in [0,1]$ such that $g(-T) neq g(T)$. Define $delta = g(T)-g(-T)$. Now for any continuous function $h$ in the $frac{delta}{3}$ ball around $g$, it’s clear that $h(T) neq h(-T)$. In other words, the entire $frac{delta}{3}$ ball around $g$ lies within $U$. So $U$ is open and therefore, $A_{+}$ is closed.
To show its interior is empty, pick any $fin A_{+}$ and consider an $varepsilon$-ball $B$ around $f$. Since this is an $varepsilon$-tube around $ f$, it is easy to see that there are continuous functions $sin B$ such that for some $T, s(T) neq s(-T)$. In other words, $A_{+}$ does not contain any open ball. So it has an empty interior.
The arguments are analogous for $A_{-}$.
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
$endgroup$
1
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
add a comment |
$begingroup$
For each of $A_{+}$ and $A_{-}$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.
First, we note what an $varepsilon$-ball around a point $f$ looks like. It is the set of all continuous functions that differ from $f$ by less than $varepsilon$ at any point in $[0,1]$. That is, the set of of all continuous functions that lie within a $varepsilon$ “tube” around $f$.
To see that the $A_{+}$ is closed, we show its complement $U$, is open. If $g in U$, then $exists T in [0,1]$ such that $g(-T) neq g(T)$. Define $delta = g(T)-g(-T)$. Now for any continuous function $h$ in the $frac{delta}{3}$ ball around $g$, it’s clear that $h(T) neq h(-T)$. In other words, the entire $frac{delta}{3}$ ball around $g$ lies within $U$. So $U$ is open and therefore, $A_{+}$ is closed.
To show its interior is empty, pick any $fin A_{+}$ and consider an $varepsilon$-ball $B$ around $f$. Since this is an $varepsilon$-tube around $ f$, it is easy to see that there are continuous functions $sin B$ such that for some $T, s(T) neq s(-T)$. In other words, $A_{+}$ does not contain any open ball. So it has an empty interior.
The arguments are analogous for $A_{-}$.
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
$endgroup$
1
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
add a comment |
$begingroup$
For each of $A_{+}$ and $A_{-}$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.
First, we note what an $varepsilon$-ball around a point $f$ looks like. It is the set of all continuous functions that differ from $f$ by less than $varepsilon$ at any point in $[0,1]$. That is, the set of of all continuous functions that lie within a $varepsilon$ “tube” around $f$.
To see that the $A_{+}$ is closed, we show its complement $U$, is open. If $g in U$, then $exists T in [0,1]$ such that $g(-T) neq g(T)$. Define $delta = g(T)-g(-T)$. Now for any continuous function $h$ in the $frac{delta}{3}$ ball around $g$, it’s clear that $h(T) neq h(-T)$. In other words, the entire $frac{delta}{3}$ ball around $g$ lies within $U$. So $U$ is open and therefore, $A_{+}$ is closed.
To show its interior is empty, pick any $fin A_{+}$ and consider an $varepsilon$-ball $B$ around $f$. Since this is an $varepsilon$-tube around $ f$, it is easy to see that there are continuous functions $sin B$ such that for some $T, s(T) neq s(-T)$. In other words, $A_{+}$ does not contain any open ball. So it has an empty interior.
The arguments are analogous for $A_{-}$.
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
$endgroup$
For each of $A_{+}$ and $A_{-}$, we can show that it is its own closure (that is, it is closed) and that its interior is empty. In other words, it is nowhere dense and therefore, immediately, a meager set.
First, we note what an $varepsilon$-ball around a point $f$ looks like. It is the set of all continuous functions that differ from $f$ by less than $varepsilon$ at any point in $[0,1]$. That is, the set of of all continuous functions that lie within a $varepsilon$ “tube” around $f$.
To see that the $A_{+}$ is closed, we show its complement $U$, is open. If $g in U$, then $exists T in [0,1]$ such that $g(-T) neq g(T)$. Define $delta = g(T)-g(-T)$. Now for any continuous function $h$ in the $frac{delta}{3}$ ball around $g$, it’s clear that $h(T) neq h(-T)$. In other words, the entire $frac{delta}{3}$ ball around $g$ lies within $U$. So $U$ is open and therefore, $A_{+}$ is closed.
To show its interior is empty, pick any $fin A_{+}$ and consider an $varepsilon$-ball $B$ around $f$. Since this is an $varepsilon$-tube around $ f$, it is easy to see that there are continuous functions $sin B$ such that for some $T, s(T) neq s(-T)$. In other words, $A_{+}$ does not contain any open ball. So it has an empty interior.
The arguments are analogous for $A_{-}$.
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
edited Dec 3 '18 at 4:57
Gaby Alfonso
697315
697315
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
answered Dec 9 '17 at 23:31
MathemagicalMathemagical
2,9241316
2,9241316
1
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
add a comment |
1
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
1
1
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
$begingroup$
Seems easier to me to use sequences to show $A+$ is closed.
$endgroup$
– zhw.
Dec 9 '17 at 23:47
add a comment |
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$begingroup$
You have to calculate the interior of the closure of these sets and show that it is empty. As a first step, can you show that $A_+$ is closed?
$endgroup$
– Mathematician 42
Dec 9 '17 at 21:03
2
$begingroup$
As mentioned in the above comment, you can show that the given sets are closed. They are also obviously subspaces. In general, you have the following: Every proper subspace of a normed vector space has empty interior.
$endgroup$
– Martin Sleziak
Dec 10 '17 at 11:20