Mixing
On cardinality: $|[0,1]|ge|{0,1}^{mathbb{N}}|$ and $|{0,1}^{mathbb{N}}|ge |(0,1)|$
$begingroup$
I have already proved the following results:
Proposition 1. Let $pinmathbb{N}$, $pge 2$.
Let $xin [0,1).$ Then exists a sequenece ${c_k}subseteq mathbb{N}$ such that $0le c_kle p-1$ for all $kinmathbb{N}$ and $$x=sum_{k=1}^{infty}frac{c_k}{p^k}.$$
$$$$
Proposition 2.
Let $pinmathbb{N}$, $pge2$. Let $xin[0,1)$ such that
begin{equation}
x=sum_{k=1}^{infty}frac{a_k}{p^k}=sum_{k=1}^{infty}frac{b_k}{p^k}
end{equation}
with two distinct sequences ${a_k}$, ${b_k}subseteqmathbb{N}$, $0le a_kle p-1$, $0le b_kle p-1$ for all $kin mathbb{N}$.
Then exists $ninmathbb{N}$ such that $a_k=0$, $b_k=p-1$ for all $kge n+1$.
Let $xin (0,1)$, then for Proposition 1, in particular, $$x=sum_{k=1}^{infty} frac{c_k}{2^k}$$ with ${c_k}in{0,1}^{mathbb{N}}$. The map $$varphi_2colon (0,1)to {0,1}^{mathbb{N}},quadvarphi(x):={c_k}$$ defined by choosing binary development with a finite number $c_kne 0$ of in ambiguous cases (see Proposition 2.), is injective. In fact let $x,yin (0,1)$, then exists ${c_k},{d_k}in{0,1}^{mathbb{N}}$ such that $$x=sum_{k=1}^{infty}frac{c_k}{2^k}quadtext{and}quad y=sum_{k=1}^{infty}frac{d_k}{2^k}.$$ We suppose that $varphi_2(x)=varphi_2(y)$, then ${c_k}={d_k}$ this means that $c_k=d_k$ for all $kinmathbb{N}$. Therefore $$x=sum_{k=1}^inftyfrac{c_k}{2^k}=sum_{k=1}^inftyfrac{d_k}{2^k}=y,$$ then $varphi_2$ is injective.
Question 1. It's correct?
Now, we cosider tha map $$varphi_3colon {0,1}^{mathbb{N}}to [0,1],quadvarphi_3({c_k}):=sum_{k=1}^inftyfrac{c_k}{3^k}$$
Question 2 How can I show that $varphi_3$ to be injective using the above proposition?
$$$$
Question 3. Is $varphi_3$ onto?
Thanks!
proof-verification elementary-set-theory proof-explanation
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
$endgroup$
add a comment |
$begingroup$
I have already proved the following results:
Proposition 1. Let $pinmathbb{N}$, $pge 2$.
Let $xin [0,1).$ Then exists a sequenece ${c_k}subseteq mathbb{N}$ such that $0le c_kle p-1$ for all $kinmathbb{N}$ and $$x=sum_{k=1}^{infty}frac{c_k}{p^k}.$$
$$$$
Proposition 2.
Let $pinmathbb{N}$, $pge2$. Let $xin[0,1)$ such that
begin{equation}
x=sum_{k=1}^{infty}frac{a_k}{p^k}=sum_{k=1}^{infty}frac{b_k}{p^k}
end{equation}
with two distinct sequences ${a_k}$, ${b_k}subseteqmathbb{N}$, $0le a_kle p-1$, $0le b_kle p-1$ for all $kin mathbb{N}$.
Then exists $ninmathbb{N}$ such that $a_k=0$, $b_k=p-1$ for all $kge n+1$.
Let $xin (0,1)$, then for Proposition 1, in particular, $$x=sum_{k=1}^{infty} frac{c_k}{2^k}$$ with ${c_k}in{0,1}^{mathbb{N}}$. The map $$varphi_2colon (0,1)to {0,1}^{mathbb{N}},quadvarphi(x):={c_k}$$ defined by choosing binary development with a finite number $c_kne 0$ of in ambiguous cases (see Proposition 2.), is injective. In fact let $x,yin (0,1)$, then exists ${c_k},{d_k}in{0,1}^{mathbb{N}}$ such that $$x=sum_{k=1}^{infty}frac{c_k}{2^k}quadtext{and}quad y=sum_{k=1}^{infty}frac{d_k}{2^k}.$$ We suppose that $varphi_2(x)=varphi_2(y)$, then ${c_k}={d_k}$ this means that $c_k=d_k$ for all $kinmathbb{N}$. Therefore $$x=sum_{k=1}^inftyfrac{c_k}{2^k}=sum_{k=1}^inftyfrac{d_k}{2^k}=y,$$ then $varphi_2$ is injective.
Question 1. It's correct?
Now, we cosider tha map $$varphi_3colon {0,1}^{mathbb{N}}to [0,1],quadvarphi_3({c_k}):=sum_{k=1}^inftyfrac{c_k}{3^k}$$
Question 2 How can I show that $varphi_3$ to be injective using the above proposition?
$$$$
Question 3. Is $varphi_3$ onto?
Thanks!
proof-verification elementary-set-theory proof-explanation
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
$endgroup$
$begingroup$
You can't choose a binary development with finitely many $c_knot = 0$. (It's like saying that all the numbers are rational). What would you take instead of the sequence $(1,0,1,0,1,0,...)$?
$endgroup$
– Yanko
Jan 17 at 20:41
$begingroup$
In the sense of the above proposition
$endgroup$
– Jack J.
Jan 17 at 20:47
$begingroup$
Right I see you only choose this binary development if there are two different ways to express $x$. This also explains how $varphi_2(x)=varphi_2(y)$ implies that $x=y$ (perhaps you want to say a word about this). Other than that it seems like the answer to Question $1$ is yes. Also the answer to Question 2 is that $varphi_3$ is not injective because if you take $a_k,b_k$ as in the proposition you will have that $varphi_3({a_k}) = varphi_3 ({b_k})$.
$endgroup$
– Yanko
Jan 17 at 20:50
$begingroup$
I do not understand. Then, in order for $varphi_2$ to be well definited, if $x$ admits two binary equal developments, by convention, I choose that which is definitively null, by virtue of the previous lemma. For my book $varphi_3$ is injjective, but I did not understand why.
$endgroup$
– Jack J.
Jan 17 at 21:20
$begingroup$
Oh right $varphi_3$ is injective (didn't realized you divide by $3^k$ instead of $2^k$) this follows from the proposition as if two expressions are equal one of the expressions must correspond to a sequence which takes the value $p-1$ in this case $p-1=2$ but we don't get such values. $varphi_3$ is not onto, it doesn't get $1$ (you can calculate the maximum by choosing all the $c_k$ to be one).
$endgroup$
– Yanko
Jan 17 at 21:54
add a comment |
$begingroup$
I have already proved the following results:
Proposition 1. Let $pinmathbb{N}$, $pge 2$.
Let $xin [0,1).$ Then exists a sequenece ${c_k}subseteq mathbb{N}$ such that $0le c_kle p-1$ for all $kinmathbb{N}$ and $$x=sum_{k=1}^{infty}frac{c_k}{p^k}.$$
$$$$
Proposition 2.
Let $pinmathbb{N}$, $pge2$. Let $xin[0,1)$ such that
begin{equation}
x=sum_{k=1}^{infty}frac{a_k}{p^k}=sum_{k=1}^{infty}frac{b_k}{p^k}
end{equation}
with two distinct sequences ${a_k}$, ${b_k}subseteqmathbb{N}$, $0le a_kle p-1$, $0le b_kle p-1$ for all $kin mathbb{N}$.
Then exists $ninmathbb{N}$ such that $a_k=0$, $b_k=p-1$ for all $kge n+1$.
Let $xin (0,1)$, then for Proposition 1, in particular, $$x=sum_{k=1}^{infty} frac{c_k}{2^k}$$ with ${c_k}in{0,1}^{mathbb{N}}$. The map $$varphi_2colon (0,1)to {0,1}^{mathbb{N}},quadvarphi(x):={c_k}$$ defined by choosing binary development with a finite number $c_kne 0$ of in ambiguous cases (see Proposition 2.), is injective. In fact let $x,yin (0,1)$, then exists ${c_k},{d_k}in{0,1}^{mathbb{N}}$ such that $$x=sum_{k=1}^{infty}frac{c_k}{2^k}quadtext{and}quad y=sum_{k=1}^{infty}frac{d_k}{2^k}.$$ We suppose that $varphi_2(x)=varphi_2(y)$, then ${c_k}={d_k}$ this means that $c_k=d_k$ for all $kinmathbb{N}$. Therefore $$x=sum_{k=1}^inftyfrac{c_k}{2^k}=sum_{k=1}^inftyfrac{d_k}{2^k}=y,$$ then $varphi_2$ is injective.
Question 1. It's correct?
Now, we cosider tha map $$varphi_3colon {0,1}^{mathbb{N}}to [0,1],quadvarphi_3({c_k}):=sum_{k=1}^inftyfrac{c_k}{3^k}$$
Question 2 How can I show that $varphi_3$ to be injective using the above proposition?
$$$$
Question 3. Is $varphi_3$ onto?
Thanks!
proof-verification elementary-set-theory proof-explanation
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
$endgroup$
I have already proved the following results:
Proposition 1. Let $pinmathbb{N}$, $pge 2$.
Let $xin [0,1).$ Then exists a sequenece ${c_k}subseteq mathbb{N}$ such that $0le c_kle p-1$ for all $kinmathbb{N}$ and $$x=sum_{k=1}^{infty}frac{c_k}{p^k}.$$
$$$$
Proposition 2.
Let $pinmathbb{N}$, $pge2$. Let $xin[0,1)$ such that
begin{equation}
x=sum_{k=1}^{infty}frac{a_k}{p^k}=sum_{k=1}^{infty}frac{b_k}{p^k}
end{equation}
with two distinct sequences ${a_k}$, ${b_k}subseteqmathbb{N}$, $0le a_kle p-1$, $0le b_kle p-1$ for all $kin mathbb{N}$.
Then exists $ninmathbb{N}$ such that $a_k=0$, $b_k=p-1$ for all $kge n+1$.
Let $xin (0,1)$, then for Proposition 1, in particular, $$x=sum_{k=1}^{infty} frac{c_k}{2^k}$$ with ${c_k}in{0,1}^{mathbb{N}}$. The map $$varphi_2colon (0,1)to {0,1}^{mathbb{N}},quadvarphi(x):={c_k}$$ defined by choosing binary development with a finite number $c_kne 0$ of in ambiguous cases (see Proposition 2.), is injective. In fact let $x,yin (0,1)$, then exists ${c_k},{d_k}in{0,1}^{mathbb{N}}$ such that $$x=sum_{k=1}^{infty}frac{c_k}{2^k}quadtext{and}quad y=sum_{k=1}^{infty}frac{d_k}{2^k}.$$ We suppose that $varphi_2(x)=varphi_2(y)$, then ${c_k}={d_k}$ this means that $c_k=d_k$ for all $kinmathbb{N}$. Therefore $$x=sum_{k=1}^inftyfrac{c_k}{2^k}=sum_{k=1}^inftyfrac{d_k}{2^k}=y,$$ then $varphi_2$ is injective.
Question 1. It's correct?
Now, we cosider tha map $$varphi_3colon {0,1}^{mathbb{N}}to [0,1],quadvarphi_3({c_k}):=sum_{k=1}^inftyfrac{c_k}{3^k}$$
Question 2 How can I show that $varphi_3$ to be injective using the above proposition?
$$$$
Question 3. Is $varphi_3$ onto?
Thanks!
proof-verification elementary-set-theory proof-explanation
proof-verification elementary-set-theory proof-explanation
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
edited Jan 18 at 3:48
Andrés E. Caicedo
65.5k8159250
65.5k8159250
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
asked Jan 17 at 20:37
Jack J.Jack J.
4292419
4292419
$begingroup$
You can't choose a binary development with finitely many $c_knot = 0$. (It's like saying that all the numbers are rational). What would you take instead of the sequence $(1,0,1,0,1,0,...)$?
$endgroup$
– Yanko
Jan 17 at 20:41
$begingroup$
In the sense of the above proposition
$endgroup$
– Jack J.
Jan 17 at 20:47
$begingroup$
Right I see you only choose this binary development if there are two different ways to express $x$. This also explains how $varphi_2(x)=varphi_2(y)$ implies that $x=y$ (perhaps you want to say a word about this). Other than that it seems like the answer to Question $1$ is yes. Also the answer to Question 2 is that $varphi_3$ is not injective because if you take $a_k,b_k$ as in the proposition you will have that $varphi_3({a_k}) = varphi_3 ({b_k})$.
$endgroup$
– Yanko
Jan 17 at 20:50
$begingroup$
I do not understand. Then, in order for $varphi_2$ to be well definited, if $x$ admits two binary equal developments, by convention, I choose that which is definitively null, by virtue of the previous lemma. For my book $varphi_3$ is injjective, but I did not understand why.
$endgroup$
– Jack J.
Jan 17 at 21:20
$begingroup$
Oh right $varphi_3$ is injective (didn't realized you divide by $3^k$ instead of $2^k$) this follows from the proposition as if two expressions are equal one of the expressions must correspond to a sequence which takes the value $p-1$ in this case $p-1=2$ but we don't get such values. $varphi_3$ is not onto, it doesn't get $1$ (you can calculate the maximum by choosing all the $c_k$ to be one).
$endgroup$
– Yanko
Jan 17 at 21:54
add a comment |
$begingroup$
You can't choose a binary development with finitely many $c_knot = 0$. (It's like saying that all the numbers are rational). What would you take instead of the sequence $(1,0,1,0,1,0,...)$?
$endgroup$
– Yanko
Jan 17 at 20:41
$begingroup$
In the sense of the above proposition
$endgroup$
– Jack J.
Jan 17 at 20:47
$begingroup$
Right I see you only choose this binary development if there are two different ways to express $x$. This also explains how $varphi_2(x)=varphi_2(y)$ implies that $x=y$ (perhaps you want to say a word about this). Other than that it seems like the answer to Question $1$ is yes. Also the answer to Question 2 is that $varphi_3$ is not injective because if you take $a_k,b_k$ as in the proposition you will have that $varphi_3({a_k}) = varphi_3 ({b_k})$.
$endgroup$
– Yanko
Jan 17 at 20:50
$begingroup$
I do not understand. Then, in order for $varphi_2$ to be well definited, if $x$ admits two binary equal developments, by convention, I choose that which is definitively null, by virtue of the previous lemma. For my book $varphi_3$ is injjective, but I did not understand why.
$endgroup$
– Jack J.
Jan 17 at 21:20
$begingroup$
Oh right $varphi_3$ is injective (didn't realized you divide by $3^k$ instead of $2^k$) this follows from the proposition as if two expressions are equal one of the expressions must correspond to a sequence which takes the value $p-1$ in this case $p-1=2$ but we don't get such values. $varphi_3$ is not onto, it doesn't get $1$ (you can calculate the maximum by choosing all the $c_k$ to be one).
$endgroup$
– Yanko
Jan 17 at 21:54
$begingroup$
You can't choose a binary development with finitely many $c_knot = 0$. (It's like saying that all the numbers are rational). What would you take instead of the sequence $(1,0,1,0,1,0,...)$?
$endgroup$
– Yanko
Jan 17 at 20:41
$begingroup$
You can't choose a binary development with finitely many $c_knot = 0$. (It's like saying that all the numbers are rational). What would you take instead of the sequence $(1,0,1,0,1,0,...)$?
$endgroup$
– Yanko
Jan 17 at 20:41
$begingroup$
In the sense of the above proposition
$endgroup$
– Jack J.
Jan 17 at 20:47
$begingroup$
In the sense of the above proposition
$endgroup$
– Jack J.
Jan 17 at 20:47
$begingroup$
Right I see you only choose this binary development if there are two different ways to express $x$. This also explains how $varphi_2(x)=varphi_2(y)$ implies that $x=y$ (perhaps you want to say a word about this). Other than that it seems like the answer to Question $1$ is yes. Also the answer to Question 2 is that $varphi_3$ is not injective because if you take $a_k,b_k$ as in the proposition you will have that $varphi_3({a_k}) = varphi_3 ({b_k})$.
$endgroup$
– Yanko
Jan 17 at 20:50
$begingroup$
Right I see you only choose this binary development if there are two different ways to express $x$. This also explains how $varphi_2(x)=varphi_2(y)$ implies that $x=y$ (perhaps you want to say a word about this). Other than that it seems like the answer to Question $1$ is yes. Also the answer to Question 2 is that $varphi_3$ is not injective because if you take $a_k,b_k$ as in the proposition you will have that $varphi_3({a_k}) = varphi_3 ({b_k})$.
$endgroup$
– Yanko
Jan 17 at 20:50
$begingroup$
I do not understand. Then, in order for $varphi_2$ to be well definited, if $x$ admits two binary equal developments, by convention, I choose that which is definitively null, by virtue of the previous lemma. For my book $varphi_3$ is injjective, but I did not understand why.
$endgroup$
– Jack J.
Jan 17 at 21:20
$begingroup$
I do not understand. Then, in order for $varphi_2$ to be well definited, if $x$ admits two binary equal developments, by convention, I choose that which is definitively null, by virtue of the previous lemma. For my book $varphi_3$ is injjective, but I did not understand why.
$endgroup$
– Jack J.
Jan 17 at 21:20
$begingroup$
Oh right $varphi_3$ is injective (didn't realized you divide by $3^k$ instead of $2^k$) this follows from the proposition as if two expressions are equal one of the expressions must correspond to a sequence which takes the value $p-1$ in this case $p-1=2$ but we don't get such values. $varphi_3$ is not onto, it doesn't get $1$ (you can calculate the maximum by choosing all the $c_k$ to be one).
$endgroup$
– Yanko
Jan 17 at 21:54
$begingroup$
Oh right $varphi_3$ is injective (didn't realized you divide by $3^k$ instead of $2^k$) this follows from the proposition as if two expressions are equal one of the expressions must correspond to a sequence which takes the value $p-1$ in this case $p-1=2$ but we don't get such values. $varphi_3$ is not onto, it doesn't get $1$ (you can calculate the maximum by choosing all the $c_k$ to be one).
$endgroup$
– Yanko
Jan 17 at 21:54
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077485%2fon-cardinality-0-1-ge-0-1-mathbbn-and-0-1-mathbbn-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077485%2fon-cardinality-0-1-ge-0-1-mathbbn-and-0-1-mathbbn-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You can't choose a binary development with finitely many $c_knot = 0$. (It's like saying that all the numbers are rational). What would you take instead of the sequence $(1,0,1,0,1,0,...)$?
$endgroup$
– Yanko
Jan 17 at 20:41
$begingroup$
In the sense of the above proposition
$endgroup$
– Jack J.
Jan 17 at 20:47
$begingroup$
Right I see you only choose this binary development if there are two different ways to express $x$. This also explains how $varphi_2(x)=varphi_2(y)$ implies that $x=y$ (perhaps you want to say a word about this). Other than that it seems like the answer to Question $1$ is yes. Also the answer to Question 2 is that $varphi_3$ is not injective because if you take $a_k,b_k$ as in the proposition you will have that $varphi_3({a_k}) = varphi_3 ({b_k})$.
$endgroup$
– Yanko
Jan 17 at 20:50
$begingroup$
I do not understand. Then, in order for $varphi_2$ to be well definited, if $x$ admits two binary equal developments, by convention, I choose that which is definitively null, by virtue of the previous lemma. For my book $varphi_3$ is injjective, but I did not understand why.
$endgroup$
– Jack J.
Jan 17 at 21:20
$begingroup$
Oh right $varphi_3$ is injective (didn't realized you divide by $3^k$ instead of $2^k$) this follows from the proposition as if two expressions are equal one of the expressions must correspond to a sequence which takes the value $p-1$ in this case $p-1=2$ but we don't get such values. $varphi_3$ is not onto, it doesn't get $1$ (you can calculate the maximum by choosing all the $c_k$ to be one).
$endgroup$
– Yanko
Jan 17 at 21:54