Two proofs involving continuity of functions? Need help?

**mfetch22** · August 12th 2010, 09:00 AM

I have two questions that I'm having trouble with, heres the first one:

[1] Suppose $f$ is defined at $a$ but not continuous at $a$ (i.e. $\lim_{x \to a} f(x) \neq f(a)$ ). Prove that for some number $\epsilon > 0$ there are numbers $x$ arbitrarily close to $a$ with $|f(x) - f(a)| > \epsilon$ . Also, give a graphical interpretation of this problem.

Heres the work I've done so far (which isn't really any work at all, just allot of scratched out pages in my note book, I'm only posting the part I don't think was a mistake within my attempts at solving):

The problem tells me that I need to prove that there are numbers arbitrarily close to $a$ , so I assume I need to show something like this:

$0 < |x-a| < \delta$

Now, I'm simply lost. I don't know which direction this proof needs to go in. I'm 100% positive that this probably seems so obvious that its silly to some of the memebers on this site, but I'm new to Analysis, so this proof stuff isn't my best area. I know how to do limit proofs, does the question involve something like a proof of a limit, in a sense?

Anyway, heres the second problem:

NOTE: This problem referances to theorems established earlier in the textbook, I will place them below so the referennces make sense:

Theorem 1

- If $f$ and $g$ are continuous at $a$ then:

(a) $f + g$ is continuous at $a$
(b) $f \cdot g$ is continuous at $a$
(c) Moreover, if $g(a) \neq 0$ , then
$\frac{1}{g}$ is continuous at $a$

Theorem 2

- If $g$ is continuous at $a$ , and $f$ is continuous at $g(a)$ , then $f(g)$ [the composition of $g$ into $f$ ] is continuous at $a$ .

Now, heres the question that referances to these theorems:

[2] Prove theorem (1)-(c) by using theorem 2 and continuity of the function $f(x) = 1/x$

I'm stumped on this one too. My main question about these questions is this:

How do you know, when a text book doesnt have many of the answers in the back, weather or not your proof is sufficiently rigourous??? I mean, consider proving limits. After the chapter in this textbook where I spent problem after problem proving limits using the epsilon delta method, is it neccesary for me to use the epsilon delta method to prove these problems about continuoity? I mean, am I allowed to use some of the theorems they showed to be true about limits? Or is it generally expected in the mathematical world Real Analysis to do every problem from the most basic theoretical and foundational things?

Thanks in advance

**Also sprach Zarathustra** · August 12th 2010, 09:11 AM

$lim_{x\to a}|f(x) - f(a)|\in \mathbb{R}^+/ \{0\}$

**mfetch22** · August 12th 2010, 09:48 AM

Originally Posted by Also sprach Zarathustra

$lim_{x\to a}|f(x) - f(a)|\in \mathbb{R}^+/ \{0\}$

I'm not completely fimmilar with the notation and conventions of set thoery. Is your post stating that:

$\lim_{x \to a} \; |f(x) - f(a)|$

is a memeber of the set of positive real numbers? Like that:

$\lim_{x \to a} \; |f(x) - f(a)| > 0$

Or does that mean something else. And I know that $\emptyset$ denotes the null set; but does $\{0\}$ mean anything different?

Oh..... I just realized my mistake in reading the notation, I think. Disragard my question about $\{0\}$ (which is just the set whos only memeber is $0$ right?). So, if I'm interpreting it correctly, your statement implies (or directly says) that:

$0 \leq \lim_{x \to a} \; |f(x) - f(a)|$

Is that a correct interpretation? (Again, I apologize, I'm not throughly educated in set theory notation and logic)

**Also sprach Zarathustra** · August 12th 2010, 10:26 AM

$lim_{x\to a}|f(x) - f(a)|\in \mathbb{R}^+/ \{0\}<br />$

Means:

$lim_{x\to a}|f(x) - f(a)|=k<br />$

when $k>0$

**Plato** · August 12th 2010, 10:37 AM

This is the definition of the statement that $f$ is continuous at $a$ :
$\left( {\forall \varepsilon > 0} \right)\left( {\exists \delta > 0} \right)\left( {\forall x} \right)\left[ {0 < \left| {x - a} \right| < \delta \to \left| {f(x) - f(a)} \right| < \varepsilon } \right]$ .

If the function is not continuous at $a$ , then negate that definition:
$\left( {\exists \varepsilon > 0} \right)\left( {\forall \delta > 0} \right)\left( {\exists y} \right)\left[ {0 < \left| {y - a} \right| < \delta \wedge \left| {f(y) - f(a)} \right| \geqslant \varepsilon } \right]$

That is the proof. Nothing more.

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