Prove that each $B_n$ is empty for $1 leq n leq n_0$












0












$begingroup$


Let $f : X to mathbb R$ be continuous and $A$ a dense subset of $X$. Let
$$B_n = {x in X| − n < f (x) < n}, n in mathbb N.$$
If for some $n_0$ the intersection $A cap B_{n_0}$ is empty, prove that each $B_n$ is empty for $1 leq n leq n_0$



Definition: A mapping $f:X mapsto Y$ is said to be continuous at the point $x$ of $X$ provided that for every $epsilon > 0$ there is a $delta >0$ such that
$$y in X, d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon$$



The mapping f is said to be continuous if it is continuous at every $x$ of $X$ .



Dense subset means $overline{A} = X $






real-analysis continuity metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    The $B_n$'s embed in each other, and, generally, if $E$ is dense and $U$ is non-empty and open, then $E cap U not = emptyset$.
    $endgroup$
    – mathworker21
    Dec 27 '18 at 22:51
















0












$begingroup$


Let $f : X to mathbb R$ be continuous and $A$ a dense subset of $X$. Let
$$B_n = {x in X| − n < f (x) < n}, n in mathbb N.$$
If for some $n_0$ the intersection $A cap B_{n_0}$ is empty, prove that each $B_n$ is empty for $1 leq n leq n_0$



Definition: A mapping $f:X mapsto Y$ is said to be continuous at the point $x$ of $X$ provided that for every $epsilon > 0$ there is a $delta >0$ such that
$$y in X, d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon$$



The mapping f is said to be continuous if it is continuous at every $x$ of $X$ .



Dense subset means $overline{A} = X $






real-analysis continuity metric-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    The $B_n$'s embed in each other, and, generally, if $E$ is dense and $U$ is non-empty and open, then $E cap U not = emptyset$.
    $endgroup$
    – mathworker21
    Dec 27 '18 at 22:51














0












0








0





$begingroup$


Let $f : X to mathbb R$ be continuous and $A$ a dense subset of $X$. Let
$$B_n = {x in X| − n < f (x) < n}, n in mathbb N.$$
If for some $n_0$ the intersection $A cap B_{n_0}$ is empty, prove that each $B_n$ is empty for $1 leq n leq n_0$



Definition: A mapping $f:X mapsto Y$ is said to be continuous at the point $x$ of $X$ provided that for every $epsilon > 0$ there is a $delta >0$ such that
$$y in X, d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon$$



The mapping f is said to be continuous if it is continuous at every $x$ of $X$ .



Dense subset means $overline{A} = X $






real-analysis continuity metric-spaces







share|cite|improve this question









$endgroup$




Let $f : X to mathbb R$ be continuous and $A$ a dense subset of $X$. Let
$$B_n = {x in X| − n < f (x) < n}, n in mathbb N.$$
If for some $n_0$ the intersection $A cap B_{n_0}$ is empty, prove that each $B_n$ is empty for $1 leq n leq n_0$



Definition: A mapping $f:X mapsto Y$ is said to be continuous at the point $x$ of $X$ provided that for every $epsilon > 0$ there is a $delta >0$ such that
$$y in X, d_X(x,y) < delta implies d_Y(f(x),f(y)) < epsilon$$



The mapping f is said to be continuous if it is continuous at every $x$ of $X$ .



Dense subset means $overline{A} = X $






real-analysis continuity metric-spaces




real-analysis continuity metric-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 27 '18 at 22:46









PumpkinPumpkin

5071419




5071419












  • $begingroup$
    The $B_n$'s embed in each other, and, generally, if $E$ is dense and $U$ is non-empty and open, then $E cap U not = emptyset$.
    $endgroup$
    – mathworker21
    Dec 27 '18 at 22:51


















  • $begingroup$
    The $B_n$'s embed in each other, and, generally, if $E$ is dense and $U$ is non-empty and open, then $E cap U not = emptyset$.
    $endgroup$
    – mathworker21
    Dec 27 '18 at 22:51
















$begingroup$
The $B_n$'s embed in each other, and, generally, if $E$ is dense and $U$ is non-empty and open, then $E cap U not = emptyset$.
$endgroup$
– mathworker21
Dec 27 '18 at 22:51




$begingroup$
The $B_n$'s embed in each other, and, generally, if $E$ is dense and $U$ is non-empty and open, then $E cap U not = emptyset$.
$endgroup$
– mathworker21
Dec 27 '18 at 22:51










1 Answer
1






active

oldest

votes


















3












$begingroup$

Clearly $B_n subseteq B_{n_0}$ for such $n$, so we need only show that $B_{n_0}$ is empty.



Suppose not. Then there is some $x in B_{n_0}$. Necessarily, $x notin A$.



Since $f$ is continuous, if we in particular take $varepsilon := n_0 - |f(x)| > 0$ (this is true since $x in B_{n_0}$), there is some $delta > 0$ such that for all $y$ in$X$ such that $d_X(x,y) < delta$, we have $d_mathbb{R}(f(x),f(y)) < varepsilon$.



But since $A$ is dense in $X$, there is, in particular, some $a in A$ such that $d_X(x,a) < delta$, hence $d_{mathbb{R}}(f(x),f(a)) < varepsilon$. But then begin{align*}|f(a)| &leq |f(x)| + |f(x)-f(a)| \&= |f(x)| + d_{mathbb{R}}(f(x),f(a)) \&< |f(x)| + varepsilon \&= |f(x)| + n_0 - |f(x)| \&= n_0,end{align*} so $a in B_{n_0}$, hence $a in Acap B_{n_0}$, but $A cap B_{n_0} = emptyset$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
    $endgroup$
    – mathworker21
    Dec 27 '18 at 23:14










  • $begingroup$
    Oops, that's what I get for editing things after too many Christmas leftovers.
    $endgroup$
    – user3482749
    Dec 27 '18 at 23:15












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

Clearly $B_n subseteq B_{n_0}$ for such $n$, so we need only show that $B_{n_0}$ is empty.



Suppose not. Then there is some $x in B_{n_0}$. Necessarily, $x notin A$.



Since $f$ is continuous, if we in particular take $varepsilon := n_0 - |f(x)| > 0$ (this is true since $x in B_{n_0}$), there is some $delta > 0$ such that for all $y$ in$X$ such that $d_X(x,y) < delta$, we have $d_mathbb{R}(f(x),f(y)) < varepsilon$.



But since $A$ is dense in $X$, there is, in particular, some $a in A$ such that $d_X(x,a) < delta$, hence $d_{mathbb{R}}(f(x),f(a)) < varepsilon$. But then begin{align*}|f(a)| &leq |f(x)| + |f(x)-f(a)| \&= |f(x)| + d_{mathbb{R}}(f(x),f(a)) \&< |f(x)| + varepsilon \&= |f(x)| + n_0 - |f(x)| \&= n_0,end{align*} so $a in B_{n_0}$, hence $a in Acap B_{n_0}$, but $A cap B_{n_0} = emptyset$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
    $endgroup$
    – mathworker21
    Dec 27 '18 at 23:14










  • $begingroup$
    Oops, that's what I get for editing things after too many Christmas leftovers.
    $endgroup$
    – user3482749
    Dec 27 '18 at 23:15
















3












$begingroup$

Clearly $B_n subseteq B_{n_0}$ for such $n$, so we need only show that $B_{n_0}$ is empty.



Suppose not. Then there is some $x in B_{n_0}$. Necessarily, $x notin A$.



Since $f$ is continuous, if we in particular take $varepsilon := n_0 - |f(x)| > 0$ (this is true since $x in B_{n_0}$), there is some $delta > 0$ such that for all $y$ in$X$ such that $d_X(x,y) < delta$, we have $d_mathbb{R}(f(x),f(y)) < varepsilon$.



But since $A$ is dense in $X$, there is, in particular, some $a in A$ such that $d_X(x,a) < delta$, hence $d_{mathbb{R}}(f(x),f(a)) < varepsilon$. But then begin{align*}|f(a)| &leq |f(x)| + |f(x)-f(a)| \&= |f(x)| + d_{mathbb{R}}(f(x),f(a)) \&< |f(x)| + varepsilon \&= |f(x)| + n_0 - |f(x)| \&= n_0,end{align*} so $a in B_{n_0}$, hence $a in Acap B_{n_0}$, but $A cap B_{n_0} = emptyset$, a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
    $endgroup$
    – mathworker21
    Dec 27 '18 at 23:14










  • $begingroup$
    Oops, that's what I get for editing things after too many Christmas leftovers.
    $endgroup$
    – user3482749
    Dec 27 '18 at 23:15














3












3








3





$begingroup$

Clearly $B_n subseteq B_{n_0}$ for such $n$, so we need only show that $B_{n_0}$ is empty.



Suppose not. Then there is some $x in B_{n_0}$. Necessarily, $x notin A$.



Since $f$ is continuous, if we in particular take $varepsilon := n_0 - |f(x)| > 0$ (this is true since $x in B_{n_0}$), there is some $delta > 0$ such that for all $y$ in$X$ such that $d_X(x,y) < delta$, we have $d_mathbb{R}(f(x),f(y)) < varepsilon$.



But since $A$ is dense in $X$, there is, in particular, some $a in A$ such that $d_X(x,a) < delta$, hence $d_{mathbb{R}}(f(x),f(a)) < varepsilon$. But then begin{align*}|f(a)| &leq |f(x)| + |f(x)-f(a)| \&= |f(x)| + d_{mathbb{R}}(f(x),f(a)) \&< |f(x)| + varepsilon \&= |f(x)| + n_0 - |f(x)| \&= n_0,end{align*} so $a in B_{n_0}$, hence $a in Acap B_{n_0}$, but $A cap B_{n_0} = emptyset$, a contradiction.






share|cite|improve this answer











$endgroup$



Clearly $B_n subseteq B_{n_0}$ for such $n$, so we need only show that $B_{n_0}$ is empty.



Suppose not. Then there is some $x in B_{n_0}$. Necessarily, $x notin A$.



Since $f$ is continuous, if we in particular take $varepsilon := n_0 - |f(x)| > 0$ (this is true since $x in B_{n_0}$), there is some $delta > 0$ such that for all $y$ in$X$ such that $d_X(x,y) < delta$, we have $d_mathbb{R}(f(x),f(y)) < varepsilon$.



But since $A$ is dense in $X$, there is, in particular, some $a in A$ such that $d_X(x,a) < delta$, hence $d_{mathbb{R}}(f(x),f(a)) < varepsilon$. But then begin{align*}|f(a)| &leq |f(x)| + |f(x)-f(a)| \&= |f(x)| + d_{mathbb{R}}(f(x),f(a)) \&< |f(x)| + varepsilon \&= |f(x)| + n_0 - |f(x)| \&= n_0,end{align*} so $a in B_{n_0}$, hence $a in Acap B_{n_0}$, but $A cap B_{n_0} = emptyset$, a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 27 '18 at 23:16

























answered Dec 27 '18 at 23:06









user3482749user3482749

4,3291119




4,3291119












  • $begingroup$
    nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
    $endgroup$
    – mathworker21
    Dec 27 '18 at 23:14










  • $begingroup$
    Oops, that's what I get for editing things after too many Christmas leftovers.
    $endgroup$
    – user3482749
    Dec 27 '18 at 23:15


















  • $begingroup$
    nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
    $endgroup$
    – mathworker21
    Dec 27 '18 at 23:14










  • $begingroup$
    Oops, that's what I get for editing things after too many Christmas leftovers.
    $endgroup$
    – user3482749
    Dec 27 '18 at 23:15
















$begingroup$
nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
$endgroup$
– mathworker21
Dec 27 '18 at 23:14




$begingroup$
nicely written. just one comment: you say "this is positive" right after 0. either say "this is true since ..." or remove the "$> 0$"
$endgroup$
– mathworker21
Dec 27 '18 at 23:14












$begingroup$
Oops, that's what I get for editing things after too many Christmas leftovers.
$endgroup$
– user3482749
Dec 27 '18 at 23:15




$begingroup$
Oops, that's what I get for editing things after too many Christmas leftovers.
$endgroup$
– user3482749
Dec 27 '18 at 23:15


















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