Mixing
Let X be a set, How to show $limlimits_{x to infty} d(x_n,y_n) = d(x,y)$ for two sequences $x_n,y_n in X$ and...
$begingroup$
Let $(X, d)$ be a metric space, let $x_n$ and $y_n$ be two sequences in $X$ and $x, y ∈ X$. Suppose that
$x_n$ converges with respect $d$ to $x$ and $y_n$ converges with respect to $d$ to $y$. Show that $limlimits_{x to infty} d(x_n,y_n) = d(x,y)$
hint : use triangle inequality several times.
I attempted this question by applying triangle inequality to the $d(x,y)$, where I get for $x,y,z in X$, $d(x,y) leq d(x,z) + d(z,y)$. but I don't see this leading anywhere.
Intuitively, I also feel like there is no need to apply the triangular inequality. Isn't it true that since $x_n to x, y_n to y$, I can write
$limlimits_{x to infty} d(x_n,y_n) = d(x,y)$ and be done with it?
Thank you for the help guys.
analysis
asked Jan 16 at 1:42
RicoRico
274
$endgroup$
add a comment |
$begingroup$
Let $(X, d)$ be a metric space, let $x_n$ and $y_n$ be two sequences in $X$ and $x, y ∈ X$. Suppose that
$x_n$ converges with respect $d$ to $x$ and $y_n$ converges with respect to $d$ to $y$. Show that $limlimits_{x to infty} d(x_n,y_n) = d(x,y)$
hint : use triangle inequality several times.
I attempted this question by applying triangle inequality to the $d(x,y)$, where I get for $x,y,z in X$, $d(x,y) leq d(x,z) + d(z,y)$. but I don't see this leading anywhere.
Intuitively, I also feel like there is no need to apply the triangular inequality. Isn't it true that since $x_n to x, y_n to y$, I can write
$limlimits_{x to infty} d(x_n,y_n) = d(x,y)$ and be done with it?
Thank you for the help guys.
analysis
asked Jan 16 at 1:42
RicoRico
274
$endgroup$
$begingroup$
This seems really simple but I am really not getting the idea to approach it. If possible please let me know if there is any hint that can be useful to me. Thanks
$endgroup$
– Rico
Jan 16 at 1:43
$begingroup$
$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$
$endgroup$
– Thomas Shelby
Jan 16 at 1:47
$begingroup$
by definition then $d(x_n,x) , d(y_n,x) = 0$, but that only shows the $leq$, do I need to show $d(x_n,y_n) geq d(x,y)$ again to show that they are equal?
$endgroup$
– Rico
Jan 16 at 1:51
add a comment |
$begingroup$
Let $(X, d)$ be a metric space, let $x_n$ and $y_n$ be two sequences in $X$ and $x, y ∈ X$. Suppose that
$x_n$ converges with respect $d$ to $x$ and $y_n$ converges with respect to $d$ to $y$. Show that $limlimits_{x to infty} d(x_n,y_n) = d(x,y)$
hint : use triangle inequality several times.
I attempted this question by applying triangle inequality to the $d(x,y)$, where I get for $x,y,z in X$, $d(x,y) leq d(x,z) + d(z,y)$. but I don't see this leading anywhere.
Intuitively, I also feel like there is no need to apply the triangular inequality. Isn't it true that since $x_n to x, y_n to y$, I can write
$limlimits_{x to infty} d(x_n,y_n) = d(x,y)$ and be done with it?
Thank you for the help guys.
analysis
asked Jan 16 at 1:42
RicoRico
274
$endgroup$
Let $(X, d)$ be a metric space, let $x_n$ and $y_n$ be two sequences in $X$ and $x, y ∈ X$. Suppose that
$x_n$ converges with respect $d$ to $x$ and $y_n$ converges with respect to $d$ to $y$. Show that $limlimits_{x to infty} d(x_n,y_n) = d(x,y)$
hint : use triangle inequality several times.
I attempted this question by applying triangle inequality to the $d(x,y)$, where I get for $x,y,z in X$, $d(x,y) leq d(x,z) + d(z,y)$. but I don't see this leading anywhere.
Intuitively, I also feel like there is no need to apply the triangular inequality. Isn't it true that since $x_n to x, y_n to y$, I can write
$limlimits_{x to infty} d(x_n,y_n) = d(x,y)$ and be done with it?
Thank you for the help guys.
analysis
analysis
asked Jan 16 at 1:42
RicoRico
274
asked Jan 16 at 1:42
RicoRico
274
asked Jan 16 at 1:42
RicoRico
274
asked Jan 16 at 1:42
RicoRico
274
asked Jan 16 at 1:42
RicoRico
274
274
$begingroup$
This seems really simple but I am really not getting the idea to approach it. If possible please let me know if there is any hint that can be useful to me. Thanks
$endgroup$
– Rico
Jan 16 at 1:43
$begingroup$
$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$
$endgroup$
– Thomas Shelby
Jan 16 at 1:47
$begingroup$
by definition then $d(x_n,x) , d(y_n,x) = 0$, but that only shows the $leq$, do I need to show $d(x_n,y_n) geq d(x,y)$ again to show that they are equal?
$endgroup$
– Rico
Jan 16 at 1:51
add a comment |
$begingroup$
This seems really simple but I am really not getting the idea to approach it. If possible please let me know if there is any hint that can be useful to me. Thanks
$endgroup$
– Rico
Jan 16 at 1:43
$begingroup$
$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$
$endgroup$
– Thomas Shelby
Jan 16 at 1:47
$begingroup$
by definition then $d(x_n,x) , d(y_n,x) = 0$, but that only shows the $leq$, do I need to show $d(x_n,y_n) geq d(x,y)$ again to show that they are equal?
$endgroup$
– Rico
Jan 16 at 1:51
$begingroup$
This seems really simple but I am really not getting the idea to approach it. If possible please let me know if there is any hint that can be useful to me. Thanks
$endgroup$
– Rico
Jan 16 at 1:43
$begingroup$
This seems really simple but I am really not getting the idea to approach it. If possible please let me know if there is any hint that can be useful to me. Thanks
$endgroup$
– Rico
Jan 16 at 1:43
$begingroup$
$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$
$endgroup$
– Thomas Shelby
Jan 16 at 1:47
$begingroup$
$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$
$endgroup$
– Thomas Shelby
Jan 16 at 1:47
$begingroup$
by definition then $d(x_n,x) , d(y_n,x) = 0$, but that only shows the $leq$, do I need to show $d(x_n,y_n) geq d(x,y)$ again to show that they are equal?
$endgroup$
– Rico
Jan 16 at 1:51
$begingroup$
by definition then $d(x_n,x) , d(y_n,x) = 0$, but that only shows the $leq$, do I need to show $d(x_n,y_n) geq d(x,y)$ again to show that they are equal?
$endgroup$
– Rico
Jan 16 at 1:51
add a comment |
1 Answer
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$begingroup$
$$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$$
and
$$d(x,y)le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$$
Now apply the limits.
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
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add a comment |
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1 Answer
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$begingroup$
$$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$$
and
$$d(x,y)le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$$
Now apply the limits.
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
add a comment |
$begingroup$
$$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$$
and
$$d(x,y)le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$$
Now apply the limits.
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
add a comment |
$begingroup$
$$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$$
and
$$d(x,y)le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$$
Now apply the limits.
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
$endgroup$
$$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$$
and
$$d(x,y)le d(x,x_n)+d(x_n,y_n)+d(y_n,y)$$
Now apply the limits.
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
answered Jan 16 at 1:59
Thomas ShelbyThomas Shelby
3,4891525
3,4891525
add a comment |
add a comment |
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$begingroup$
This seems really simple but I am really not getting the idea to approach it. If possible please let me know if there is any hint that can be useful to me. Thanks
$endgroup$
– Rico
Jan 16 at 1:43
$begingroup$
$d(x_n,y_n)le d(x_n,x)+d(x,y)+d(y,y_n)$
$endgroup$
– Thomas Shelby
Jan 16 at 1:47
$begingroup$
by definition then $d(x_n,x) , d(y_n,x) = 0$, but that only shows the $leq$, do I need to show $d(x_n,y_n) geq d(x,y)$ again to show that they are equal?
$endgroup$
– Rico
Jan 16 at 1:51