Are My Definitions of Continuity and Uniform Continuity Correct?

[mtvateallmybrains]

I'm seeking a bit of affirmation or correction here before i try to solidify this to memory...

I know continuity to mean:
Let f:D -> R (D being an interval we know to be the domain, D)
Let x_0 be a member of the domain, D.
This implies that the function f is continuous at the point x_0 iff
for any e >0 there exists a d>0 such that x belongs to the domain, D AND |x-x_0|< d => |f(x)-f(x_0)| < e .

I interpret this to mean:
This is the criterion by which we judge if some function (f) is continuous at whatever-point-we-wish-to-test-for-continuity-at (x_0) over some interval that is, in the least, a subset of the domain (if not the entire domain itself).

////////

I know uniform continuity to mean:
Let a compact set, K be a subset of R. Let f:K->R. Then f is uniformly continuous on the set K.

I interpret this to mean:
The previous definition of continuity is now applicable to any and every point that is a member of the compact set, K. In other words, the interval/set over which K is defined satisfies the previous criterion of continuity at all points in K.

====
Is there a need to adjust either my definition (as quoted by my prof. for an introductory advanced calculus class) or my interpretation of these concepts - or are they within a reasonable tolerance of "precise-ness" for the _actual_ definition/interpretation/distinction of the concept of continuity and of the concept of uniform continuity? Please advise, thank you!

 

[mathman]

A simple way to look at it is in the definition of d. In the definition of continuity, d may depend on x. For uniform continuity, d is independent of x. In both cases, d will depend on e.

 

[mtvateallmybrains]

so in the case of continuity we select x, and in the case of uniform continuity we select delta in terms of epsilon?
========================
Also, it would probably be most helpful (at least to myself, if not others) if someone were to respond to the first post as if it were four True/False statements: (just like the old days)
IF a statement is True, THEN please say that "Yes, that statement (in it's entirety) is necessarily and sufficiently True." ELSE, the statement is FALSE.
If the statement is False, please explain why (what necessary and sufficient conditions - while also taking into considering the introductory nature/level of this material - were wrong and/or missing & what, if anything, is extraneous?).

This will hopefully eliminate a great deal of confusion and the potential for ambiguity. Also note that I'm not asking for an explanation here (unless what I have asserted is FALSE), I'm asking for an affirmation.

Thanks!

 

[NateTG]

Right:
We say that a function is continuous on an interval $D$ if for any $x_0$ on the interval, and for any $\epsilon > 0$ there exists [tex]\delta > 0[/itex] with the property that $x \in D$ and $| x_0 - x| < \delta$ implies that $|f(x)-f(x_0)| < \epsilon$<br /> <br /> A function is unformly continuous on an interval $D$ if for any $\epsilon>0$ there exists $\delta > 0$ so that for any $$x_1,x_2 \in D$$, $|x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)| < \epsilon$<br /> <br /> For example $$f(x)=\frac{1}{x}$$ is contiuous, but not uniformly continuous on the interval $$(0,+\infty)$$.[/tex]

 

[matt grime]

You don't actually define uniform continuity, so it's hard to say if you are correct or not in that.

A function that is continuous on a compact set is uniformly continuous, yes, but that isn't the definition.

Basically, a function is continuous at a point x in the domain if... etc ... *where d depends on both e and x*, so your definition is correct (I'm not at all sure what your interpretation is an interpretation of, though).

it is uniformly continuous if d can be chosen such that there is no dependence on x.

 

[mathman]

To NateTG: In your definition of uniform continuity, you switched epsilon and delta in the final implication.

 

[NateTG]

To NateTG: In your definition of uniform continuity, you switched epsilon and delta in the final implication.

Oops. (fixing)