1. ## Continuity practice

thanks all

2. We know that $\left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right|$.
Using continuity make each of the two on the right $< \frac{\varepsilon }{2}$.

3. Hm, so $\left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right| \leqslant 2\epsilon$.

So, $\left| {f(x) - f(c)} \right|\leqslant \frac{\epsilon}{2}$

So $\delta= \frac{\epsilon}{2}$

Then $2|f(x)-f(y)|<2\frac{\epsilon}{2}$

How is this?

Also,
Originally Posted by Plato
We know that $\left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right|$.
Is that triangle inequality? How did you recognize that you could do this? What happens when you substitute c for y?

Thanks

4. Originally Posted by sfspitfire23
Hm, so $\left| {f(x) - f(y)} \right| \leqslant \left| {f(x) - f(c)} \right| + \left| {f(x) - f(c)} \right| \leqslant 2\epsilon$.

So, $\left| {f(x) - f(c)} \right|\leqslant \frac{\epsilon}{2}$

So $\delta= \frac{\epsilon}{2}$

Then $2|f(x)-f(y)|<2\frac{\epsilon}{2}$

How is this?
I don't get the reference to $\delta=\frac{\varepsilon}{2}$. I think you mean choose $\delta$ so that $d(f(c),f(x))<\frac{\varepsilon}{2}$. Otherwise looks fine.

Also,

Is that triangle inequality? How did you recognize that you could do this? What happens when you substitute c for y?

Thanks
Yes this is the triangle inequlaity and Plato made a slight typo, it should be $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|$

5. 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 $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$. I feel like choosing $2\epsilon$ 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 $2\epsilon$, I chose delta so that $|f(x)-f(c)|<\frac{\epsilon}{2}$ and $|f(y)-f(c)|< \frac{\epsilon}{2}$. Then I can take $2|f(x)-f(y)|$ because this will still be $\leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But then we can say that $|f(x)-f(y)|\leqslant \frac{\epsilon}{2}$ because really, $y$ is just some $c$. (THIS is the part that I'm not sure on. Is y just some c?) Then we get $2|f(x)-f(y)|<2\frac{\epsilon}{2}$

6. Originally Posted by sfspitfire23
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 $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$. I feel like choosing $2\epsilon$ 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 $2\epsilon$, I chose delta so that $|f(x)-f(c)|<\frac{\epsilon}{2}$ and $|f(y)-f(c)|< \frac{\epsilon}{2}$. Then I can take $2|f(x)-f(y)|$ because this will still be $\leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But then we can say that $|f(x)-f(y)|\leqslant \frac{\epsilon}{2}$ because really, $y$ is just some $c$. (THIS is the part that I'm not sure on. Is y just some c?) Then we get $2|f(x)-f(y)|<2\frac{\epsilon}{2}$

Try rewriting it with this in mind.

$|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But, since $f$ is continuous at $c$ you can make those last two terms arbitrarily small by choosing $x,y$ arbitrarily close to $c$...

7. Ok, trying again, please tell me what you think

We know that $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$. Because $f$ is continuous at $c$, I can make $x, y$ arbitrarily small so that $|f(x)-f(c)|<\epsilon$ and $|f(y)-f(c)|<\epsilon$
Now, choosing a delta to get rid of the 2 in the $2\epsilon$, I chose delta so that $|f(x)-f(c)|<\frac{\epsilon}{2}$ and $|f(y)-f(c)|< \frac{\epsilon}{2}$. Then I can take $2|f(x)-f(y)|$ because this will still be $\leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But then we can say that $|f(x)-f(y)|\leqslant \frac{\epsilon}{2}$ because really, $y$ is just some $c$. (THIS is the part that I'm not sure on. Is y just some c?) Then we get $2|f(x)-f(y)|<2\frac{\epsilon}{2}$[/quote]

8. Originally Posted by sfspitfire23
Ok, trying again, please tell me what you think

We know that $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$. Because $f$ is continuous at $c$, I can make $x, y$ arbitrarily small so that $|f(x)-f(c)|<\epsilon$ and $|f(y)-f(c)|<\epsilon$
Now, choosing a delta to get rid of the 2 in the $2\epsilon$, I chose delta so that $|f(x)-f(c)|<\frac{\epsilon}{2}$ and $|f(y)-f(c)|< \frac{\epsilon}{2}$. Then I can take $2|f(x)-f(y)|$ because this will still be $\leqslant |f(x)-f(c)|+|f(y)-f(c)|$. But then we can say that $|f(x)-f(y)|\leqslant \frac{\epsilon}{2}$ because really, $y$ is just some $c$. (THIS is the part that I'm not sure on. Is y just some c?) Then we get $2|f(x)-f(y)|<2\frac{\epsilon}{2}$
[/QUOTE]
You are messing up the order in which you say things.

The fact that $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)|<
\varepsilon$
is the solution to the problem. You have to note that you can do this since you can make $|f(x)-f(c)|$ and $|f(y)-f(c)|$ by taking $\delta=\min\{\delta_1,\delta_2\}$ where $|x-c|<\delta_1\implies |f(x)-f(c)|<\varepsilon$ and $|y-c|<\delta_2\implies |f(y)-f(c)|<\varepsilon$

9. So, we have $|f(x)-f(y)|\leqslant |f(x)-f(c)|+|f(y)-f(c)| \leqslant 2\epsilon$ because we can find some delta1/delta2 such that

$|f(x)-f(c)|<\epsilon$ and $|f(y)-f(c)|<\epsilon$

Choose $\delta = \epsilon /2$

But then $2|f(x)-f(y)|<|f(x)-f(c)|+|f(y)-f(c)|$

So $2|f(x)-f(y)|\Rightarrow 2|f(x)-f(c)|$ because y is really just some c and we then have $2|f(x)-f(c)|<2(\epsilon / 2)$