thanks all
Hm, so $\displaystyle \left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right| \leqslant 2\epsilon$.
So, $\displaystyle \left| {f(x) - f(c)} \right|\leqslant \frac{\epsilon}{2}$
So $\displaystyle \delta= \frac{\epsilon}{2}$
Then $\displaystyle 2|f(x)-f(y)|<2\frac{\epsilon}{2}$
How is this?
Also, Is that triangle inequality? How did you recognize that you could do this? What happens when you substitute c for y?
Thanks
I don't get the reference to $\displaystyle \delta=\frac{\varepsilon}{2}$. I think you mean choose $\displaystyle \delta$ so that $\displaystyle d(f(c),f(x))<\frac{\varepsilon}{2}$. Otherwise looks fine.
Yes this is the triangle inequlaity and Plato made a slight typo, it should be $\displaystyle |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|$Also,
Is that triangle inequality? How did you recognize that you could do this? What happens when you substitute c for y?
Thanks
So heres what I have, I'm not sure about my logic at one step (I point it out where it is)
We know that $\displaystyle |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$. I feel like choosing $\displaystyle 2\epsilon$ because I feel like it will be east to pick my delta to cancel the 2 (is this logic correct?). Now, choosing a delta to get rid of the 2 in the $\displaystyle 2\epsilon$, I chose delta so that $\displaystyle |f(x)-f(c)|<\frac{\epsilon}{2}$ and $\displaystyle |f(y)-f(c)|< \frac{\epsilon}{2}$. Then I can take $\displaystyle 2|f(x)-f(y)|$ because this will still be $\displaystyle \leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But then we can say that $\displaystyle |f(x)-f(y)|\leqslant \frac{\epsilon}{2}$ because really, $\displaystyle y$ is just some $\displaystyle c$. (THIS is the part that I'm not sure on. Is y just some c?) Then we get $\displaystyle 2|f(x)-f(y)|<2\frac{\epsilon}{2}$
I can't really follow your logic.
Try rewriting it with this in mind.
$\displaystyle |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But, since $\displaystyle f$ is continuous at $\displaystyle c$ you can make those last two terms arbitrarily small by choosing $\displaystyle x,y$ arbitrarily close to $\displaystyle c$...
Ok, trying again, please tell me what you think
We know that $\displaystyle |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$. Because $\displaystyle f$ is continuous at $\displaystyle c$, I can make $\displaystyle x, y$ arbitrarily small so that $\displaystyle |f(x)-f(c)|<\epsilon$ and $\displaystyle |f(y)-f(c)|<\epsilon$
Now, choosing a delta to get rid of the 2 in the $\displaystyle 2\epsilon$, I chose delta so that $\displaystyle |f(x)-f(c)|<\frac{\epsilon}{2}$ and $\displaystyle |f(y)-f(c)|< \frac{\epsilon}{2}$. Then I can take $\displaystyle 2|f(x)-f(y)|$ because this will still be $\displaystyle \leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But then we can say that $\displaystyle |f(x)-f(y)|\leqslant \frac{\epsilon}{2}$ because really, $\displaystyle y$ is just some $\displaystyle c$. (THIS is the part that I'm not sure on. Is y just some c?) Then we get $\displaystyle 2|f(x)-f(y)|<2\frac{\epsilon}{2}$[/quote]
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You are messing up the order in which you say things.
The fact that $\displaystyle |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|<
\varepsilon$ is the solution to the problem. You have to note that you can do this since you can make $\displaystyle |f(x)-f(c)|$ and $\displaystyle |f(y)-f(c)|$ by taking $\displaystyle \delta=\min\{\delta_1,\delta_2\}$ where $\displaystyle |x-c|<\delta_1\implies |f(x)-f(c)|<\varepsilon$ and $\displaystyle |y-c|<\delta_2\implies |f(y)-f(c)|<\varepsilon$
So, we have $\displaystyle |f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$ because we can find some delta1/delta2 such that
$\displaystyle |f(x)-f(c)|<\epsilon$ and $\displaystyle |f(y)-f(c)|<\epsilon$
Choose $\displaystyle \delta = \epsilon /2$
But then $\displaystyle 2|f(x)-f(y)|<|f(x)-f(c)|+|f(y)-f(c)|$
So $\displaystyle 2|f(x)-f(y)|\Rightarrow 2|f(x)-f(c)|$ because y is really just some c and we then have $\displaystyle 2|f(x)-f(c)|<2(\epsilon / 2)$