Mixing
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Proof that the multiplicaiton $mathbbRtimes mathbbR to mathbbR$ is continuous
Clash Royale CLAN TAG#URR8PPP
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I'm having troubles understanding a proof of the continuity of the product function $cdot:mathbbRtimes mathbbR to mathbbR$. The proof starts by considering a point $(x_0,y_0) = z_0$, and by letting $M$ to be the greatest between the two numbers $|x_0|+1$ and $|y_0|+1$. Given
$epsilon gt 0$, we let $delta = min1,fracepsilon2M$. Calling $W$ the neighborhood $[x_0-delta,x_0+delta]times [y_0-delta,y_0+delta]$ centered at he point $(x_0,y_0)$ of $mathbbRtimes mathbbR$ of radius $delta$, we should notice that, if $(x,y)in W$, $|y|leq |y_0|+delta leq |y_0|+1$: then we can write: $$|xy-x_0y_0| leq |x-x_0||y|+|x_0||y-y_0| leq (|x-x_0|+|y-y_0|)M leq dots leq epsilon$$
I'm wondering why should be $|y|leq |y_0|+delta leq |y_0|+1$ and how the guy who wrote that proof choose $M$ and $delta$ to be precisely those values. (I've seen a lot of times, in trying to find an explanaiton, techniques that lies upon choices that looks similar to these, in proving results about continuity in topological vector spaces).
Continuing with the proof, another thing that doesn't convince me is the conclusion that states: "Therefore, for every $(x,y)in W$, their product $xy$ belongs to a neighborhood of $z_0$ of radius $epsilon$", but I hope this is a printing mistake, and that the author wanted to write, at the beginning, $z_0 = x_0y_0$.
real-analysis continuity proof-explanation
asked yesterday
marco21
9517
add a comment |Â
up vote
1
down vote
favorite
I'm having troubles understanding a proof of the continuity of the product function $cdot:mathbbRtimes mathbbR to mathbbR$. The proof starts by considering a point $(x_0,y_0) = z_0$, and by letting $M$ to be the greatest between the two numbers $|x_0|+1$ and $|y_0|+1$. Given
$epsilon gt 0$, we let $delta = min1,fracepsilon2M$. Calling $W$ the neighborhood $[x_0-delta,x_0+delta]times [y_0-delta,y_0+delta]$ centered at he point $(x_0,y_0)$ of $mathbbRtimes mathbbR$ of radius $delta$, we should notice that, if $(x,y)in W$, $|y|leq |y_0|+delta leq |y_0|+1$: then we can write: $$|xy-x_0y_0| leq |x-x_0||y|+|x_0||y-y_0| leq (|x-x_0|+|y-y_0|)M leq dots leq epsilon$$
I'm wondering why should be $|y|leq |y_0|+delta leq |y_0|+1$ and how the guy who wrote that proof choose $M$ and $delta$ to be precisely those values. (I've seen a lot of times, in trying to find an explanaiton, techniques that lies upon choices that looks similar to these, in proving results about continuity in topological vector spaces).
Continuing with the proof, another thing that doesn't convince me is the conclusion that states: "Therefore, for every $(x,y)in W$, their product $xy$ belongs to a neighborhood of $z_0$ of radius $epsilon$", but I hope this is a printing mistake, and that the author wanted to write, at the beginning, $z_0 = x_0y_0$.
real-analysis continuity proof-explanation
asked yesterday
marco21
9517
I'm not sure I'm following. In the first bit of your second paragraph ("I'm wondering...") are you asking how we know that abs(y) <= abs(y_0) <= abs(y_0)+1, or are you asking why those values were chosen to highlight in the first place?
– nitsua60
yesterday
I'm asking for a proof (or a hint of a proof) of that: I've not been able to check it. (Excuse my English)
– marco21
yesterday
1
Regarding your why question: $|y|$ is at most the larger of $|y_0-delta|$ and $|y_0+delta|$. Both of which are no greater than $|y_0| + delta$. The next equality follows from the fact that $delta le 1$ by definition. As for how, I’ll leave to someone else...
– Theoretical Economist
yesterday
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having troubles understanding a proof of the continuity of the product function $cdot:mathbbRtimes mathbbR to mathbbR$. The proof starts by considering a point $(x_0,y_0) = z_0$, and by letting $M$ to be the greatest between the two numbers $|x_0|+1$ and $|y_0|+1$. Given
$epsilon gt 0$, we let $delta = min1,fracepsilon2M$. Calling $W$ the neighborhood $[x_0-delta,x_0+delta]times [y_0-delta,y_0+delta]$ centered at he point $(x_0,y_0)$ of $mathbbRtimes mathbbR$ of radius $delta$, we should notice that, if $(x,y)in W$, $|y|leq |y_0|+delta leq |y_0|+1$: then we can write: $$|xy-x_0y_0| leq |x-x_0||y|+|x_0||y-y_0| leq (|x-x_0|+|y-y_0|)M leq dots leq epsilon$$
I'm wondering why should be $|y|leq |y_0|+delta leq |y_0|+1$ and how the guy who wrote that proof choose $M$ and $delta$ to be precisely those values. (I've seen a lot of times, in trying to find an explanaiton, techniques that lies upon choices that looks similar to these, in proving results about continuity in topological vector spaces).
Continuing with the proof, another thing that doesn't convince me is the conclusion that states: "Therefore, for every $(x,y)in W$, their product $xy$ belongs to a neighborhood of $z_0$ of radius $epsilon$", but I hope this is a printing mistake, and that the author wanted to write, at the beginning, $z_0 = x_0y_0$.
real-analysis continuity proof-explanation
asked yesterday
marco21
9517
I'm having troubles understanding a proof of the continuity of the product function $cdot:mathbbRtimes mathbbR to mathbbR$. The proof starts by considering a point $(x_0,y_0) = z_0$, and by letting $M$ to be the greatest between the two numbers $|x_0|+1$ and $|y_0|+1$. Given
$epsilon gt 0$, we let $delta = min1,fracepsilon2M$. Calling $W$ the neighborhood $[x_0-delta,x_0+delta]times [y_0-delta,y_0+delta]$ centered at he point $(x_0,y_0)$ of $mathbbRtimes mathbbR$ of radius $delta$, we should notice that, if $(x,y)in W$, $|y|leq |y_0|+delta leq |y_0|+1$: then we can write: $$|xy-x_0y_0| leq |x-x_0||y|+|x_0||y-y_0| leq (|x-x_0|+|y-y_0|)M leq dots leq epsilon$$
I'm wondering why should be $|y|leq |y_0|+delta leq |y_0|+1$ and how the guy who wrote that proof choose $M$ and $delta$ to be precisely those values. (I've seen a lot of times, in trying to find an explanaiton, techniques that lies upon choices that looks similar to these, in proving results about continuity in topological vector spaces).
Continuing with the proof, another thing that doesn't convince me is the conclusion that states: "Therefore, for every $(x,y)in W$, their product $xy$ belongs to a neighborhood of $z_0$ of radius $epsilon$", but I hope this is a printing mistake, and that the author wanted to write, at the beginning, $z_0 = x_0y_0$.
real-analysis continuity proof-explanation
asked yesterday
marco21
9517
edited yesterday
asked yesterday
marco21
9517
asked yesterday
marco21
9517
asked yesterday
marco21
9517
9517
I'm not sure I'm following. In the first bit of your second paragraph ("I'm wondering...") are you asking how we know that abs(y) <= abs(y_0) <= abs(y_0)+1, or are you asking why those values were chosen to highlight in the first place?
– nitsua60
yesterday
I'm asking for a proof (or a hint of a proof) of that: I've not been able to check it. (Excuse my English)
– marco21
yesterday
1
Regarding your why question: $|y|$ is at most the larger of $|y_0-delta|$ and $|y_0+delta|$. Both of which are no greater than $|y_0| + delta$. The next equality follows from the fact that $delta le 1$ by definition. As for how, I’ll leave to someone else...
– Theoretical Economist
yesterday
add a comment |Â
I'm not sure I'm following. In the first bit of your second paragraph ("I'm wondering...") are you asking how we know that abs(y) <= abs(y_0) <= abs(y_0)+1, or are you asking why those values were chosen to highlight in the first place?
– nitsua60
yesterday
I'm asking for a proof (or a hint of a proof) of that: I've not been able to check it. (Excuse my English)
– marco21
yesterday
1
Regarding your why question: $|y|$ is at most the larger of $|y_0-delta|$ and $|y_0+delta|$. Both of which are no greater than $|y_0| + delta$. The next equality follows from the fact that $delta le 1$ by definition. As for how, I’ll leave to someone else...
– Theoretical Economist
yesterday
I'm not sure I'm following. In the first bit of your second paragraph ("I'm wondering...") are you asking how we know that abs(y) <= abs(y_0) <= abs(y_0)+1, or are you asking why those values were chosen to highlight in the first place?
– nitsua60
yesterday
I'm not sure I'm following. In the first bit of your second paragraph ("I'm wondering...") are you asking how we know that abs(y) <= abs(y_0) <= abs(y_0)+1, or are you asking why those values were chosen to highlight in the first place?
– nitsua60
yesterday
I'm asking for a proof (or a hint of a proof) of that: I've not been able to check it. (Excuse my English)
– marco21
yesterday
I'm asking for a proof (or a hint of a proof) of that: I've not been able to check it. (Excuse my English)
– marco21
yesterday
1
1
Regarding your why question: $|y|$ is at most the larger of $|y_0-delta|$ and $|y_0+delta|$. Both of which are no greater than $|y_0| + delta$. The next equality follows from the fact that $delta le 1$ by definition. As for how, I’ll leave to someone else...
– Theoretical Economist
yesterday
Regarding your why question: $|y|$ is at most the larger of $|y_0-delta|$ and $|y_0+delta|$. Both of which are no greater than $|y_0| + delta$. The next equality follows from the fact that $delta le 1$ by definition. As for how, I’ll leave to someone else...
– Theoretical Economist
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The best way to understand how or why the author made that specific choice for $delta$ and $M$ a priori, is to commence a proof using only the information at hand and see what develops. As you will see, other choices are possible and need not be determined at the start by some flash of insight.
The objective is to find $delta > 0$ such that for all $(x,y) in W$ we have $|xy - x_0y_0| < epsilon.$
You have to anticipate that the determination of a suitable $delta$ will require some estimation and bounding. The tools we have immediately at our disposal are $$(x,y) in [x_0-delta,x_0 + delta] times [y_0-delta,y_0 + delta],$$ which implies
$$|x - x_0| leqslant delta, quad |y - y_0| leqslant delta,$$
along with,
$$quad |x| leqslant |x_0| + |x - x_0| leqslant |x_0| + delta, \ ,,,,|y| leqslant |y_0| + |y - y_0| leqslant ,,|y_0| + ,delta $$
The last two lines follow from the reverse triangle inequality or, if you prefer, the usual triangle inequality with $|x| = |x_0 + (x - x_0) | leqslant |x_0| + |x-x_0|$.
Since we have only simple inequalities involving $x$ and $y$ separately, we make the decomposition $|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0|$ and then obtain using those inequalities,
$$tag*|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0| leqslant delta(|y_0| + delta) + delta|x_0|\ leqslant 2delta max(|x_0|,|y_0| + delta)$$
We want to find $delta$ such that the RHS of (*) is smaller than $epsilon$, but at this point the procedure will be messy. When $|y_0| > |x_0|$ we are left with a quadratic equation, and to examine the other case where $|x_0| > |y_0| + delta$ we are faced with an iterative process.
However, if we are not concerned with finding the largest possible $delta$, then we can set a constraint $delta < alpha$ where $alpha > 0$ is arbitrary and replace $max(|x_0|, |y_0| + delta)$ with $M = max(|x_0|, |y_0| + alpha)$, to obtain
$$|xy- x_0y_0| leqslant 2delta M$$
If both $delta leqslant alpha$ and $delta leqslant epsilon/(2M)$, that is $delta leqslant min(alpha,epsilon/(2M))$, then $|xy-x_0y_0| leqslant epsilon$.
The author happened to like another choice, $M = max(|x_0|+1,|y_0| + 1)$, perhaps because it looks "cleaner", and knew this would work in advance at the start of the proof.
Finally, you are correct in that the statement should be $xy$ belongs to a neighborhood of $x_0y_0$ of radius $epsilon$. Here neighborhood means closed ball with center $x_0y_0$ and radius $epsilon$.
answered yesterday
RRL
43.3k42160
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The best way to understand how or why the author made that specific choice for $delta$ and $M$ a priori, is to commence a proof using only the information at hand and see what develops. As you will see, other choices are possible and need not be determined at the start by some flash of insight.
The objective is to find $delta > 0$ such that for all $(x,y) in W$ we have $|xy - x_0y_0| < epsilon.$
You have to anticipate that the determination of a suitable $delta$ will require some estimation and bounding. The tools we have immediately at our disposal are $$(x,y) in [x_0-delta,x_0 + delta] times [y_0-delta,y_0 + delta],$$ which implies
$$|x - x_0| leqslant delta, quad |y - y_0| leqslant delta,$$
along with,
$$quad |x| leqslant |x_0| + |x - x_0| leqslant |x_0| + delta, \ ,,,,|y| leqslant |y_0| + |y - y_0| leqslant ,,|y_0| + ,delta $$
The last two lines follow from the reverse triangle inequality or, if you prefer, the usual triangle inequality with $|x| = |x_0 + (x - x_0) | leqslant |x_0| + |x-x_0|$.
Since we have only simple inequalities involving $x$ and $y$ separately, we make the decomposition $|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0|$ and then obtain using those inequalities,
$$tag*|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0| leqslant delta(|y_0| + delta) + delta|x_0|\ leqslant 2delta max(|x_0|,|y_0| + delta)$$
We want to find $delta$ such that the RHS of (*) is smaller than $epsilon$, but at this point the procedure will be messy. When $|y_0| > |x_0|$ we are left with a quadratic equation, and to examine the other case where $|x_0| > |y_0| + delta$ we are faced with an iterative process.
However, if we are not concerned with finding the largest possible $delta$, then we can set a constraint $delta < alpha$ where $alpha > 0$ is arbitrary and replace $max(|x_0|, |y_0| + delta)$ with $M = max(|x_0|, |y_0| + alpha)$, to obtain
$$|xy- x_0y_0| leqslant 2delta M$$
If both $delta leqslant alpha$ and $delta leqslant epsilon/(2M)$, that is $delta leqslant min(alpha,epsilon/(2M))$, then $|xy-x_0y_0| leqslant epsilon$.
The author happened to like another choice, $M = max(|x_0|+1,|y_0| + 1)$, perhaps because it looks "cleaner", and knew this would work in advance at the start of the proof.
Finally, you are correct in that the statement should be $xy$ belongs to a neighborhood of $x_0y_0$ of radius $epsilon$. Here neighborhood means closed ball with center $x_0y_0$ and radius $epsilon$.
answered yesterday
RRL
43.3k42160
add a comment |Â
up vote
2
down vote
accepted
The best way to understand how or why the author made that specific choice for $delta$ and $M$ a priori, is to commence a proof using only the information at hand and see what develops. As you will see, other choices are possible and need not be determined at the start by some flash of insight.
The objective is to find $delta > 0$ such that for all $(x,y) in W$ we have $|xy - x_0y_0| < epsilon.$
You have to anticipate that the determination of a suitable $delta$ will require some estimation and bounding. The tools we have immediately at our disposal are $$(x,y) in [x_0-delta,x_0 + delta] times [y_0-delta,y_0 + delta],$$ which implies
$$|x - x_0| leqslant delta, quad |y - y_0| leqslant delta,$$
along with,
$$quad |x| leqslant |x_0| + |x - x_0| leqslant |x_0| + delta, \ ,,,,|y| leqslant |y_0| + |y - y_0| leqslant ,,|y_0| + ,delta $$
The last two lines follow from the reverse triangle inequality or, if you prefer, the usual triangle inequality with $|x| = |x_0 + (x - x_0) | leqslant |x_0| + |x-x_0|$.
Since we have only simple inequalities involving $x$ and $y$ separately, we make the decomposition $|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0|$ and then obtain using those inequalities,
$$tag*|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0| leqslant delta(|y_0| + delta) + delta|x_0|\ leqslant 2delta max(|x_0|,|y_0| + delta)$$
We want to find $delta$ such that the RHS of (*) is smaller than $epsilon$, but at this point the procedure will be messy. When $|y_0| > |x_0|$ we are left with a quadratic equation, and to examine the other case where $|x_0| > |y_0| + delta$ we are faced with an iterative process.
However, if we are not concerned with finding the largest possible $delta$, then we can set a constraint $delta < alpha$ where $alpha > 0$ is arbitrary and replace $max(|x_0|, |y_0| + delta)$ with $M = max(|x_0|, |y_0| + alpha)$, to obtain
$$|xy- x_0y_0| leqslant 2delta M$$
If both $delta leqslant alpha$ and $delta leqslant epsilon/(2M)$, that is $delta leqslant min(alpha,epsilon/(2M))$, then $|xy-x_0y_0| leqslant epsilon$.
The author happened to like another choice, $M = max(|x_0|+1,|y_0| + 1)$, perhaps because it looks "cleaner", and knew this would work in advance at the start of the proof.
Finally, you are correct in that the statement should be $xy$ belongs to a neighborhood of $x_0y_0$ of radius $epsilon$. Here neighborhood means closed ball with center $x_0y_0$ and radius $epsilon$.
answered yesterday
RRL
43.3k42160
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The best way to understand how or why the author made that specific choice for $delta$ and $M$ a priori, is to commence a proof using only the information at hand and see what develops. As you will see, other choices are possible and need not be determined at the start by some flash of insight.
The objective is to find $delta > 0$ such that for all $(x,y) in W$ we have $|xy - x_0y_0| < epsilon.$
You have to anticipate that the determination of a suitable $delta$ will require some estimation and bounding. The tools we have immediately at our disposal are $$(x,y) in [x_0-delta,x_0 + delta] times [y_0-delta,y_0 + delta],$$ which implies
$$|x - x_0| leqslant delta, quad |y - y_0| leqslant delta,$$
along with,
$$quad |x| leqslant |x_0| + |x - x_0| leqslant |x_0| + delta, \ ,,,,|y| leqslant |y_0| + |y - y_0| leqslant ,,|y_0| + ,delta $$
The last two lines follow from the reverse triangle inequality or, if you prefer, the usual triangle inequality with $|x| = |x_0 + (x - x_0) | leqslant |x_0| + |x-x_0|$.
Since we have only simple inequalities involving $x$ and $y$ separately, we make the decomposition $|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0|$ and then obtain using those inequalities,
$$tag*|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0| leqslant delta(|y_0| + delta) + delta|x_0|\ leqslant 2delta max(|x_0|,|y_0| + delta)$$
We want to find $delta$ such that the RHS of (*) is smaller than $epsilon$, but at this point the procedure will be messy. When $|y_0| > |x_0|$ we are left with a quadratic equation, and to examine the other case where $|x_0| > |y_0| + delta$ we are faced with an iterative process.
However, if we are not concerned with finding the largest possible $delta$, then we can set a constraint $delta < alpha$ where $alpha > 0$ is arbitrary and replace $max(|x_0|, |y_0| + delta)$ with $M = max(|x_0|, |y_0| + alpha)$, to obtain
$$|xy- x_0y_0| leqslant 2delta M$$
If both $delta leqslant alpha$ and $delta leqslant epsilon/(2M)$, that is $delta leqslant min(alpha,epsilon/(2M))$, then $|xy-x_0y_0| leqslant epsilon$.
The author happened to like another choice, $M = max(|x_0|+1,|y_0| + 1)$, perhaps because it looks "cleaner", and knew this would work in advance at the start of the proof.
Finally, you are correct in that the statement should be $xy$ belongs to a neighborhood of $x_0y_0$ of radius $epsilon$. Here neighborhood means closed ball with center $x_0y_0$ and radius $epsilon$.
answered yesterday
RRL
43.3k42160
The best way to understand how or why the author made that specific choice for $delta$ and $M$ a priori, is to commence a proof using only the information at hand and see what develops. As you will see, other choices are possible and need not be determined at the start by some flash of insight.
The objective is to find $delta > 0$ such that for all $(x,y) in W$ we have $|xy - x_0y_0| < epsilon.$
You have to anticipate that the determination of a suitable $delta$ will require some estimation and bounding. The tools we have immediately at our disposal are $$(x,y) in [x_0-delta,x_0 + delta] times [y_0-delta,y_0 + delta],$$ which implies
$$|x - x_0| leqslant delta, quad |y - y_0| leqslant delta,$$
along with,
$$quad |x| leqslant |x_0| + |x - x_0| leqslant |x_0| + delta, \ ,,,,|y| leqslant |y_0| + |y - y_0| leqslant ,,|y_0| + ,delta $$
The last two lines follow from the reverse triangle inequality or, if you prefer, the usual triangle inequality with $|x| = |x_0 + (x - x_0) | leqslant |x_0| + |x-x_0|$.
Since we have only simple inequalities involving $x$ and $y$ separately, we make the decomposition $|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0|$ and then obtain using those inequalities,
$$tag*|xy- x_0y_0| leqslant |x - x_0||y| + |x_0||y - y_0| leqslant delta(|y_0| + delta) + delta|x_0|\ leqslant 2delta max(|x_0|,|y_0| + delta)$$
We want to find $delta$ such that the RHS of (*) is smaller than $epsilon$, but at this point the procedure will be messy. When $|y_0| > |x_0|$ we are left with a quadratic equation, and to examine the other case where $|x_0| > |y_0| + delta$ we are faced with an iterative process.
However, if we are not concerned with finding the largest possible $delta$, then we can set a constraint $delta < alpha$ where $alpha > 0$ is arbitrary and replace $max(|x_0|, |y_0| + delta)$ with $M = max(|x_0|, |y_0| + alpha)$, to obtain
$$|xy- x_0y_0| leqslant 2delta M$$
If both $delta leqslant alpha$ and $delta leqslant epsilon/(2M)$, that is $delta leqslant min(alpha,epsilon/(2M))$, then $|xy-x_0y_0| leqslant epsilon$.
The author happened to like another choice, $M = max(|x_0|+1,|y_0| + 1)$, perhaps because it looks "cleaner", and knew this would work in advance at the start of the proof.
Finally, you are correct in that the statement should be $xy$ belongs to a neighborhood of $x_0y_0$ of radius $epsilon$. Here neighborhood means closed ball with center $x_0y_0$ and radius $epsilon$.
answered yesterday
RRL
43.3k42160
edited yesterday
answered yesterday
RRL
43.3k42160
answered yesterday
RRL
43.3k42160
answered yesterday
RRL
43.3k42160
43.3k42160
add a comment |Â
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I'm not sure I'm following. In the first bit of your second paragraph ("I'm wondering...") are you asking how we know that abs(y) <= abs(y_0) <= abs(y_0)+1, or are you asking why those values were chosen to highlight in the first place?
– nitsua60
yesterday
I'm asking for a proof (or a hint of a proof) of that: I've not been able to check it. (Excuse my English)
– marco21
yesterday
1
Regarding your why question: $|y|$ is at most the larger of $|y_0-delta|$ and $|y_0+delta|$. Both of which are no greater than $|y_0| + delta$. The next equality follows from the fact that $delta le 1$ by definition. As for how, I’ll leave to someone else...
– Theoretical Economist
yesterday