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Show a subspace is closed

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Poirot

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Feb 15, 2012
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Let L^2 be the usual vector space of complex sequences.

Let F be the subspace of sequences whose first term is zero. Show that F is closed.

Let $((V_{nk}):k=1,2,.....)$ be a convergent sequence in F. I need to show it converges to a sequence whose first term is 0. Well, for all positive integers n, we have


$V_{nk}->x_{n}$ as k tends to infinity. In particular, $V_{n1}->x_{1}$, so I need to show x_{1}=0. Problem is, I don't see why this should be true. I know $V_{11}=1$ but individual terms are of no consequence in the convergence of a sequence.
 

Opalg

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Feb 7, 2012
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Let L^2 be the usual vector space of complex sequences.

Let F be the subspace of sequences whose first term is zero. Show that F is closed.

Let $((V_{nk}):k=1,2,.....)$ be a convergent sequence in F. I need to show it converges to a sequence whose first term is 0. Well, for all positive integers n, we have


$V_{nk}->x_{n}$ as k tends to infinity. In particular, $V_{n1}->x_{1}$, so I need to show x_{1}=0. Problem is, I don't see why this should be true. I know $V_{11}=1$ but individual terms are of no consequence in the convergence of a sequence.
$L^2$-convergence implies coordinatewise convergence. The reason is that $\|x\|^2 = \sum |x_n|^2$, and it follows that if $\|x\|<\varepsilon$ then $|x_n|<\varepsilon$ for all $n$.
 
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Poirot

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Feb 15, 2012
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$L^2$-convergence implies coordinatewise convergence. The reason is that $\|x\|^2 = \sum |x_n|^2$, and it follows that if $\|x\|<\varepsilon$ then $|x_n|<\varepsilon$ for all $n$.
can you please elucidate on what you mean by co-ordinate wise convergence?

Are you saying that if I let $(x_{n})=((V_{n_{k}}):\{k=1,2...\})$ and $(y_{k})$ the limit, we have $||x_{n}-y_{n}||^2= \sum |x_{k}-y{k}|^2$ These subscripts are rather confusing.
 
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Opalg

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Feb 7, 2012
2,725
can you please elucidate on what you mean by co-ordinate wise convergence?

Are you saying that if I let $(x_{n})=((V_{nk}:k=1,2...)$ and $(y_{k})$ the limit, we have $||x_{n}-y_{n}||^2= \sum |x_{k}-y{k}|^2$ These subscripts are rather confusing.
Maybe the subscripts will be less confusing if we make one of them a superscript. Suppose you write your sequence as $\{V_n^{(k)}:k=1,2,\ldots\}$, so that $V_n^{(k)}$ means the $n$th coordinate of the $k$th vector.

The question says that $F$ is "the subspace of sequences whose first term is zero." That is a very misleading form of words. I am quite sure that it does not mean that the first member of the sequence is the zero vector. What it should say is that $F$ is the subspace consisting of all sequences whose first coordinate is zero. In other words, it is not saying that $V^{(1)}_n = 0$ for all $n$, but that $V^{(k)}_1 = 0$ for all $k$.

My previous hint was intended to say that if $V^{(k)}\to x$ in $L^2$ (as $k\to \infty$) then $V^{(k)}_n\to x_n$ (as $k\to \infty$) for each $n$ (and in particular for $n=1$).
 
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Poirot

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Feb 15, 2012
250
Maybe the subscripts will be less confusing if we make one of them a superscript. Suppose you write your sequence as $\{V_n^{(k)}:k=1,2,\ldots\}$, so that $V_n^{(k)}$ means the $n$th coordinate of the $k$th vector.

The question says that $F$ is "the subspace of sequences whose first term is zero." That is a very misleading form of words. I am quite sure that it does not mean that the first member of the sequence is the zero vector. What it should say is that $F$ is the subspace consisting of all sequences whose first coordinate is zero. In other words, it is not saying that $V^{(1)}_n = 0$ for all $n$, but that $V^{(k)}_1 = 0$ for all $k$.

My previous hint was intended to say that if $V^{(k)}\to x$ in $L^2$ (as $k\to \infty$) then $V^{(k)}_n\to x_n$ (as $k\to \infty$) for each $n$ (and in particular for $n=1$).
It seems you have confused my n and k. I was using them in the opposite roles to which you have given them. So I need to show $x_{1}=0$. So I need to show convergence in the complex numbers?
 

Opalg

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Feb 7, 2012
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It seems you have confused my n and k. I was using them in the opposite roles to which you have given them. So I need to show $x_{1}=0$. So I need to show convergence in the complex numbers?
In your initial post, you wrote "$V_{nk}\to x_{n}$ as k tends to infinity. In particular, $V_{n1}\to x_{1}$". You need to sort out which of those subscript refers to the vector and which one refers to the coordinate. There is only one vector $x$, so when you wrote $x_n$ I assumed that you meant $n$ to refer to the coordinate. But the statement "$V_{n1}\to x_{1}$" looks very confused to me, because the $k$ in $V_{nk}$ has become a $1$, whereas the $n$ in $x_n$ has become a $1$. If $k$ is tending to infinity then the $n$ should stay as an $n$, not become a $1$. That is why I suggested making one of the subscripts into a superscript. That would help you to keep the two suffixes from becoming confused.
 
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Poirot

Banned
Feb 15, 2012
250
In your initial post, you wrote "$V_{nk}\to x_{n}$ as k tends to infinity. In particular, $V_{n1}\to x_{1}$". You need to sort out which of those subscript refers to the vector and which one refers to the coordinate. There is only one vector $x$, so when you wrote $x_n$ I assumed that you meant $n$ to refer to the coordinate. But the statement "$V_{n1}\to x_{1}$" looks very confused to me, because the $k$ in $V_{nk}$ has become a $1$, whereas the $n$ in $x_n$ has become a $1$. If $k$ is tending to infinity then the $n$ should stay as an $n$, not become a $1$. That is why I suggested making one of the subscripts into a superscript. That would help you to keep the two suffixes from becoming confused.
Ok I will say it in words to help us both. We have a sequence of sequences, tending to a sequence, say x. I posited (and I think you backed this up) that the 'first' sequence tends to to the first term of x. To show x is in F, I need to show that the first term of x is 0; i.e the first sequence tends to 0 in the complex numbers. Beyond that I am stuck. Thanks for your patience
 

Opalg

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Feb 7, 2012
2,725
Ok I will say it in words to help us both. We have a sequence of sequences, tending to a sequence, say x. I posited (and I think you backed this up) that the 'first' sequence tends to to the first term of x. To show x is in F, I need to show that the first term of x is 0; i.e the first sequence tends to 0 in the complex numbers. Beyond that I am stuck. Thanks for your patience
We're still not getting very far with this. I'll have one more attempt to clear up the confusion, and if that doesn't work I'll have to leave it to someone else.

Ok, let's try saying it in words. We have a sequence of vectors. Each vector consists of a sequence of coordinates. The vectors are tending to a limit vector, say $x$. That implies that the first coordinates of the vectors tend to the first coordinate of $x$. Each vector in the sequence is in $F$. That means that the first coordinate of each vector is $0$. But a sequence of $0$s tends to $0$. Therefore the first coordinate of $x$ is $0$ and hence $x$ is in $F$.