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Proof that the sum of two continuous maps $Xto V$ is again continuous
The Next CEO of Stack OverflowHow is the metric topology the coarsest to make the metric function continuous?Continuity of the sum of continuous functionsAxiomatizing topology through continuous mapsShowing that $f$ continuousContinuity of the maximum of finite continuous functionsContinuous maps and compact-open topologyIs there another topology on $mathbbR$ that gives the same continuous functions from $mathbbR$ to $mathbbR$?Definition a weakly continuous mappingInducing differentiable structure via continuous mapsAlternative characterizations of the strong topology in a normed linear space
$begingroup$
Let $(X,mathcalO$) be a topological space, $(V,| cdot |)$ be a normed vector space, $mathcalO_V$ be the topology induced by $| cdot |$ on $V$.
I want to show, that if $f,g:X to V$ are two continuous functions, then $f+g:Xto V$ is again continuous.
The definition I'm working with is the following: $f$ is called continuous iff all preimages of open sets are again open.
A long time ago I've seen proofs for the special case $V=mathbbK^n$ and where $mathcalO$ is the topology induced by a metric $d$ on $X$, but couldn't generalize them.
I tried to show this directly by the definition, but got stuck. Here is what I tried:
Let $S=B_r(v) mid rgeq 0, v in V$. Since $S$ is subbase of $mathcalO_V$, it suffices to show $(f+g)^-1(B)in mathcalO$ for all $Bin S$.
So, let $B_r(v)in S$. We have $$ xin (f+g)^-1(B_r(v)) iff | f(x)+g(x)-v| < r$$
But this is not really helpful, since $g(x)$ can basically be everything.
general-topology continuity
asked yesterday
Jakob B.Jakob B.
518111
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcalO$) be a topological space, $(V,| cdot |)$ be a normed vector space, $mathcalO_V$ be the topology induced by $| cdot |$ on $V$.
I want to show, that if $f,g:X to V$ are two continuous functions, then $f+g:Xto V$ is again continuous.
The definition I'm working with is the following: $f$ is called continuous iff all preimages of open sets are again open.
A long time ago I've seen proofs for the special case $V=mathbbK^n$ and where $mathcalO$ is the topology induced by a metric $d$ on $X$, but couldn't generalize them.
I tried to show this directly by the definition, but got stuck. Here is what I tried:
Let $S=B_r(v) mid rgeq 0, v in V$. Since $S$ is subbase of $mathcalO_V$, it suffices to show $(f+g)^-1(B)in mathcalO$ for all $Bin S$.
So, let $B_r(v)in S$. We have $$ xin (f+g)^-1(B_r(v)) iff | f(x)+g(x)-v| < r$$
But this is not really helpful, since $g(x)$ can basically be everything.
general-topology continuity
asked yesterday
Jakob B.Jakob B.
518111
$endgroup$
add a comment |
$begingroup$
Let $(X,mathcalO$) be a topological space, $(V,| cdot |)$ be a normed vector space, $mathcalO_V$ be the topology induced by $| cdot |$ on $V$.
I want to show, that if $f,g:X to V$ are two continuous functions, then $f+g:Xto V$ is again continuous.
The definition I'm working with is the following: $f$ is called continuous iff all preimages of open sets are again open.
A long time ago I've seen proofs for the special case $V=mathbbK^n$ and where $mathcalO$ is the topology induced by a metric $d$ on $X$, but couldn't generalize them.
I tried to show this directly by the definition, but got stuck. Here is what I tried:
Let $S=B_r(v) mid rgeq 0, v in V$. Since $S$ is subbase of $mathcalO_V$, it suffices to show $(f+g)^-1(B)in mathcalO$ for all $Bin S$.
So, let $B_r(v)in S$. We have $$ xin (f+g)^-1(B_r(v)) iff | f(x)+g(x)-v| < r$$
But this is not really helpful, since $g(x)$ can basically be everything.
general-topology continuity
asked yesterday
Jakob B.Jakob B.
518111
$endgroup$
Let $(X,mathcalO$) be a topological space, $(V,| cdot |)$ be a normed vector space, $mathcalO_V$ be the topology induced by $| cdot |$ on $V$.
I want to show, that if $f,g:X to V$ are two continuous functions, then $f+g:Xto V$ is again continuous.
The definition I'm working with is the following: $f$ is called continuous iff all preimages of open sets are again open.
A long time ago I've seen proofs for the special case $V=mathbbK^n$ and where $mathcalO$ is the topology induced by a metric $d$ on $X$, but couldn't generalize them.
I tried to show this directly by the definition, but got stuck. Here is what I tried:
Let $S=B_r(v) mid rgeq 0, v in V$. Since $S$ is subbase of $mathcalO_V$, it suffices to show $(f+g)^-1(B)in mathcalO$ for all $Bin S$.
So, let $B_r(v)in S$. We have $$ xin (f+g)^-1(B_r(v)) iff | f(x)+g(x)-v| < r$$
But this is not really helpful, since $g(x)$ can basically be everything.
general-topology continuity
general-topology continuity
asked yesterday
Jakob B.Jakob B.
518111
asked yesterday
Jakob B.Jakob B.
518111
asked yesterday
Jakob B.Jakob B.
518111
asked yesterday
Jakob B.Jakob B.
518111
asked yesterday
Jakob B.Jakob B.
518111
518111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $U$ be open in $V$. Let $x_0 in (f+g)^-1(U)$. Then $f(x_0)+g(x_0) in U$, so there exists $r>0$ such that $B(f(x_0)+g(x_0),r) subset U$. Consider $V=f^-1(B(f(x_0),r/2) cap g^-1(B(f(x_0),r/2)$. This set is open in $X$ because it is an intersection of two open sets. Let $xin V$. Now can you use triangle inequality for the norm to show that $x_0 in V subset (f+g)^-1(U)$?
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Let $U$ be open in $V$. Let $x_0 in (f+g)^-1(U)$. Then $f(x_0)+g(x_0) in U$, so there exists $r>0$ such that $B(f(x_0)+g(x_0),r) subset U$. Consider $V=f^-1(B(f(x_0),r/2) cap g^-1(B(f(x_0),r/2)$. This set is open in $X$ because it is an intersection of two open sets. Let $xin V$. Now can you use triangle inequality for the norm to show that $x_0 in V subset (f+g)^-1(U)$?
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
add a comment |
$begingroup$
Let $U$ be open in $V$. Let $x_0 in (f+g)^-1(U)$. Then $f(x_0)+g(x_0) in U$, so there exists $r>0$ such that $B(f(x_0)+g(x_0),r) subset U$. Consider $V=f^-1(B(f(x_0),r/2) cap g^-1(B(f(x_0),r/2)$. This set is open in $X$ because it is an intersection of two open sets. Let $xin V$. Now can you use triangle inequality for the norm to show that $x_0 in V subset (f+g)^-1(U)$?
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
add a comment |
$begingroup$
Let $U$ be open in $V$. Let $x_0 in (f+g)^-1(U)$. Then $f(x_0)+g(x_0) in U$, so there exists $r>0$ such that $B(f(x_0)+g(x_0),r) subset U$. Consider $V=f^-1(B(f(x_0),r/2) cap g^-1(B(f(x_0),r/2)$. This set is open in $X$ because it is an intersection of two open sets. Let $xin V$. Now can you use triangle inequality for the norm to show that $x_0 in V subset (f+g)^-1(U)$?
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
$endgroup$
Let $U$ be open in $V$. Let $x_0 in (f+g)^-1(U)$. Then $f(x_0)+g(x_0) in U$, so there exists $r>0$ such that $B(f(x_0)+g(x_0),r) subset U$. Consider $V=f^-1(B(f(x_0),r/2) cap g^-1(B(f(x_0),r/2)$. This set is open in $X$ because it is an intersection of two open sets. Let $xin V$. Now can you use triangle inequality for the norm to show that $x_0 in V subset (f+g)^-1(U)$?
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
answered yesterday
Kavi Rama MurthyKavi Rama Murthy
70.8k53170
70.8k53170
add a comment |
add a comment |
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