Help with proof: If |x - x0|, |y - y0| < epsilon/2, then

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jwpaine

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My Professor has helped me map out a course of Spivak study which I will be working on independently of my calculus sequence, to help me gain a better understanding of the material.

Here is the problem I am working on from the chapter on basic properties of numbers.

Prove that if

\(\displaystyle \L |x - x_0| \,< \frac{\epsilon}{2} \,\,and\,\, |y-y_0| \,< \frac{\epsilon}{2}\)

then

\(\displaystyle \L |(x + y) - (x_o + y_0)| \,<\, \epsilon,\)
\(\displaystyle \L |(x - y) - (x_o - y_0)| \,<\, \epsilon\)

Can someone give me a good hint? I'm not sure how I would go about proving this due to my lack of understanding.
 
Here is #2. Be sure that you see where I used the triangle inequality.
\(\displaystyle \L \begin{array}{rcl}
\left| {\left( {x - y} \right) - \left( {x_0 - y_0 } \right)} \right| & = & \left| {x - x_0 - y + y_0 } \right| \\
& = & \left| {\left( {x - x_0 } \right) - \left( {y - y_0 } \right)} \right| \\
& \le & \left| {\left( {x - x_0 } \right)} \right| + \left| {\left( {-y + y_0 } \right)} \right| \\
& = & \left| {\left( {x - x_0 } \right)} \right| + \left| {\left( {y - y_0 } \right)} \right| \\
& < & \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon \\
\end{array}\)
 
pka said:
Here is #2. Be sure that you see where I used the triangle inequality.
\(\displaystyle \L \begin{array}{rcl}
\left| {\left( {x - y} \right) - \left( {x_0 - y_0 } \right)} \right| & = & \left| {x - x_0 - y + y_0 } \right| \\
& = & \left| {\left( {x - x_0 } \right) - \left( {y - y_0 } \right)} \right| \\
& \le & \left| {\left( {x - x_0 } \right)} \right| + \left| {\left( {-y + y_0 } \right)} \right| \\
& = & \left| {\left( {x - x_0 } \right)} \right| + \left| {\left( {y - y_0 } \right)} \right| \\
& < & \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon \\
\end{array}\)
Click to expand...

Why that triangle inequality on the third one up from the bottom? (less than or equal to) how did it become that?
 
\(\displaystyle \L \left| {x - x_0 - y + y_0 } \right| = \left| {\left( {x - x_0 } \right) + \left( { - y + y_0 } \right)} \right| \le \left| {\left( {x - x_0 } \right)} \right| + \left| {\left( { - y + y_0 } \right)} \right|\)

Is that what you are asking about?
You do understand the triangle inequality? Don't you?

The triangle inequality is the basis for all these sorts of problems.
 
\(\displaystyle \L |x \,+\, y| \,\,\,<=\,\,\, |x| \,+\, |y|\)

I see now: thanks.
 
OK - I was making this out to be harder than it is...

so for the first one, because of the triangle inequality:

|(x + y) - (xo + yo)|

= |x - xo + y + yo|

= |(x - xo) + (y + yo)|

<= |(x - xo)| + |(y - yo)|

< (e/2)2

< e

Right?
 
Isn't the y<sub>0</sub> suppose to be negative in your second and third line? And then you fixed it up afterward so I'm guessing it was just a typo. Looks good to me.
 
o_O said:
Isn't the y<sub>0</sub> suppose to be negative in your second and third line? And then you fixed it up afterward so I'm guessing it was just a typo. Looks good to me.
Click to expand...

Not sure what you mean... can you elaborate?

if I start with |(x + y) - (xo + yo)| than that = |x + y - xo + yo| before I group.

I then said |x + y - xo + yo| = |x - xo + y + yo| = |(x - xo) + (y + yo)|

I guess I made a mistake, BECAUSE |(y + yo)| does not equal |(y - yo)| so my proof for the first inequality doesn't work
 
If you start out with |(x+y) - (x<sub>0</sub> + y<sub>0</sub>)|, then the negative sign applies to each term in (x<sub>0</sub> + y<sub>0</sub>) so you get:

|x + y - x<sub>0</sub> - y<sub>0</sub>| = |(x - x<sub>0</sub>) + (y - y<sub>0</sub>)|

Just a minor mistake, that's all. Other than that, everything else follows.
 
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