Finite-dimensional subspaces of normed linear spaces are closed: Do there exist alternative proofs?
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It is well-known that each finite-dimensional subspace $F$ of a (real or complex) normed linear space $(V,lVert - rVert)$ is closed. The standard proof is this.
$lVert - rVert$ induces a norm $lVert - rVert_F$ on $F$. Since all norms on $F$ are equivalent, $(F,lVert - rVert_F)$ is complete so that $F$ is closed in $(V,lVert - rVert)$.
Let us now consider a finite-dimensional $V$ and a codimension $1$ subspace $F$. I tried to find a "direct proof" that $F$ is closed in $V$, but all my attempts were in vain. By a direct proof I mean one based on elementary properties of norms (essentially the triangle inequality).
Question: Does there exist a proof not using the "heavy gun" that all norms on finite-dimensional linear spaces are equivalent?
functional-analysis normed-spaces
asked Aug 21 at 8:46
Paul Frost
4,548423
 |Â
show 5 more comments
up vote
0
down vote
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It is well-known that each finite-dimensional subspace $F$ of a (real or complex) normed linear space $(V,lVert - rVert)$ is closed. The standard proof is this.
$lVert - rVert$ induces a norm $lVert - rVert_F$ on $F$. Since all norms on $F$ are equivalent, $(F,lVert - rVert_F)$ is complete so that $F$ is closed in $(V,lVert - rVert)$.
Let us now consider a finite-dimensional $V$ and a codimension $1$ subspace $F$. I tried to find a "direct proof" that $F$ is closed in $V$, but all my attempts were in vain. By a direct proof I mean one based on elementary properties of norms (essentially the triangle inequality).
Question: Does there exist a proof not using the "heavy gun" that all norms on finite-dimensional linear spaces are equivalent?
functional-analysis normed-spaces
asked Aug 21 at 8:46
Paul Frost
4,548423
You may use the continuity of linear forms.
– Michael Hoppe
Aug 21 at 8:51
@MichaelHoppe Okay, that would require the Hahn-Banach theorem. Or are there more elementary proofs for the continuity of linear functionals?
– Paul Frost
Aug 21 at 8:56
Aren’t linear maps in normed and finite dimensional vector spaces continuous per se?
– Michael Hoppe
Aug 21 at 9:00
If I may ask, what's the motivation ? (I've never heard the "all norms are equivalent" argument being referred to as a "heavy guns" argument)
– Max
Aug 21 at 9:05
1
No, you can take a sequence of linear combinations of $v_i$'s, look at the coefficients as a sequence of vectors in Euclidean space and use Bolzano-Wierstrass to prove continuity. Only definition of norm is required for this proof.
– Kavi Rama Murthy
Aug 21 at 9:28
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
It is well-known that each finite-dimensional subspace $F$ of a (real or complex) normed linear space $(V,lVert - rVert)$ is closed. The standard proof is this.
$lVert - rVert$ induces a norm $lVert - rVert_F$ on $F$. Since all norms on $F$ are equivalent, $(F,lVert - rVert_F)$ is complete so that $F$ is closed in $(V,lVert - rVert)$.
Let us now consider a finite-dimensional $V$ and a codimension $1$ subspace $F$. I tried to find a "direct proof" that $F$ is closed in $V$, but all my attempts were in vain. By a direct proof I mean one based on elementary properties of norms (essentially the triangle inequality).
Question: Does there exist a proof not using the "heavy gun" that all norms on finite-dimensional linear spaces are equivalent?
functional-analysis normed-spaces
asked Aug 21 at 8:46
Paul Frost
4,548423
It is well-known that each finite-dimensional subspace $F$ of a (real or complex) normed linear space $(V,lVert - rVert)$ is closed. The standard proof is this.
$lVert - rVert$ induces a norm $lVert - rVert_F$ on $F$. Since all norms on $F$ are equivalent, $(F,lVert - rVert_F)$ is complete so that $F$ is closed in $(V,lVert - rVert)$.
Let us now consider a finite-dimensional $V$ and a codimension $1$ subspace $F$. I tried to find a "direct proof" that $F$ is closed in $V$, but all my attempts were in vain. By a direct proof I mean one based on elementary properties of norms (essentially the triangle inequality).
Question: Does there exist a proof not using the "heavy gun" that all norms on finite-dimensional linear spaces are equivalent?
functional-analysis normed-spaces
asked Aug 21 at 8:46
Paul Frost
4,548423
asked Aug 21 at 8:46
Paul Frost
4,548423
asked Aug 21 at 8:46
Paul Frost
4,548423
asked Aug 21 at 8:46
Paul Frost
4,548423
4,548423
You may use the continuity of linear forms.
– Michael Hoppe
Aug 21 at 8:51
@MichaelHoppe Okay, that would require the Hahn-Banach theorem. Or are there more elementary proofs for the continuity of linear functionals?
– Paul Frost
Aug 21 at 8:56
Aren’t linear maps in normed and finite dimensional vector spaces continuous per se?
– Michael Hoppe
Aug 21 at 9:00
If I may ask, what's the motivation ? (I've never heard the "all norms are equivalent" argument being referred to as a "heavy guns" argument)
– Max
Aug 21 at 9:05
1
No, you can take a sequence of linear combinations of $v_i$'s, look at the coefficients as a sequence of vectors in Euclidean space and use Bolzano-Wierstrass to prove continuity. Only definition of norm is required for this proof.
– Kavi Rama Murthy
Aug 21 at 9:28
 |Â
show 5 more comments
You may use the continuity of linear forms.
– Michael Hoppe
Aug 21 at 8:51
@MichaelHoppe Okay, that would require the Hahn-Banach theorem. Or are there more elementary proofs for the continuity of linear functionals?
– Paul Frost
Aug 21 at 8:56
Aren’t linear maps in normed and finite dimensional vector spaces continuous per se?
– Michael Hoppe
Aug 21 at 9:00
If I may ask, what's the motivation ? (I've never heard the "all norms are equivalent" argument being referred to as a "heavy guns" argument)
– Max
Aug 21 at 9:05
1
No, you can take a sequence of linear combinations of $v_i$'s, look at the coefficients as a sequence of vectors in Euclidean space and use Bolzano-Wierstrass to prove continuity. Only definition of norm is required for this proof.
– Kavi Rama Murthy
Aug 21 at 9:28
You may use the continuity of linear forms.
– Michael Hoppe
Aug 21 at 8:51
You may use the continuity of linear forms.
– Michael Hoppe
Aug 21 at 8:51
@MichaelHoppe Okay, that would require the Hahn-Banach theorem. Or are there more elementary proofs for the continuity of linear functionals?
– Paul Frost
Aug 21 at 8:56
@MichaelHoppe Okay, that would require the Hahn-Banach theorem. Or are there more elementary proofs for the continuity of linear functionals?
– Paul Frost
Aug 21 at 8:56
Aren’t linear maps in normed and finite dimensional vector spaces continuous per se?
– Michael Hoppe
Aug 21 at 9:00
Aren’t linear maps in normed and finite dimensional vector spaces continuous per se?
– Michael Hoppe
Aug 21 at 9:00
If I may ask, what's the motivation ? (I've never heard the "all norms are equivalent" argument being referred to as a "heavy guns" argument)
– Max
Aug 21 at 9:05
If I may ask, what's the motivation ? (I've never heard the "all norms are equivalent" argument being referred to as a "heavy guns" argument)
– Max
Aug 21 at 9:05
1
1
No, you can take a sequence of linear combinations of $v_i$'s, look at the coefficients as a sequence of vectors in Euclidean space and use Bolzano-Wierstrass to prove continuity. Only definition of norm is required for this proof.
– Kavi Rama Murthy
Aug 21 at 9:28
No, you can take a sequence of linear combinations of $v_i$'s, look at the coefficients as a sequence of vectors in Euclidean space and use Bolzano-Wierstrass to prove continuity. Only definition of norm is required for this proof.
– Kavi Rama Murthy
Aug 21 at 9:28
 |Â
show 5 more comments
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You may use the continuity of linear forms.
– Michael Hoppe
Aug 21 at 8:51
@MichaelHoppe Okay, that would require the Hahn-Banach theorem. Or are there more elementary proofs for the continuity of linear functionals?
– Paul Frost
Aug 21 at 8:56
Aren’t linear maps in normed and finite dimensional vector spaces continuous per se?
– Michael Hoppe
Aug 21 at 9:00
If I may ask, what's the motivation ? (I've never heard the "all norms are equivalent" argument being referred to as a "heavy guns" argument)
– Max
Aug 21 at 9:05
1
No, you can take a sequence of linear combinations of $v_i$'s, look at the coefficients as a sequence of vectors in Euclidean space and use Bolzano-Wierstrass to prove continuity. Only definition of norm is required for this proof.
– Kavi Rama Murthy
Aug 21 at 9:28