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Proof that the sum of two continuous maps $Xto V$ is again continuous

1

1

$begingroup$

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

share|cite|improve this question

asked yesterday

Jakob B.Jakob B.

518111

$endgroup$

add a comment | 

1

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

asked yesterday

Jakob B.Jakob B.

518111

$endgroup$

share|cite|improve this question

asked yesterday

Jakob B.Jakob B.

518111

share|cite|improve this question

asked yesterday

Jakob B.Jakob B.

518111

asked yesterday

Jakob B.Jakob B.

518111

asked yesterday

Jakob B.Jakob B.

518111

1 Answer
1

active

oldest

votes

0

$begingroup$

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 $x_0 in V subset (f+g)^-1(U)$?

share|cite|improve this answer

answered yesterday

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

$endgroup$

add a comment | 

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0

$begingroup$

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 $x_0 in V subset (f+g)^-1(U)$?

share|cite|improve this answer

answered yesterday

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

$endgroup$

add a comment | 

0

$begingroup$

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 $x_0 in V subset (f+g)^-1(U)$?

share|cite|improve this answer

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 $x_0 in V subset (f+g)^-1(U)$?

share|cite|improve this answer

answered yesterday

Kavi Rama MurthyKavi Rama Murthy

70.8k53170

$endgroup$

share|cite|improve this answer

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