Continuity of a function

In summary: Anyway, the above is stating that if you take a real number and approximate it by rational numbers, then the limit of this sequence is the real number.
  • #1
ELESSAR TELKONT
44
0
My problem is this. Let [tex]f:\mathbb{R}^{2}\longrightarrow \mathbb{R}^{2}[/tex] be a continuous function that satifies that [tex]\forall q\in\mathbb{Q}\times\mathbb{Q}[/tex] we have [tex]f(q)=q[/tex]. Proof that [tex]\forall x\in\mathbb{R}^{2}[/tex] we have [tex]f(x)=x[/tex].

I have worked out that because it is continuous, [tex]f[/tex] satisfies that
[tex]\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(a)\longleftrightarrow f(x)\in B_{\epsilon}(f(a))[/tex]

and then [tex]\forall q\in\mathbb{Q}\times\mathbb{Q}[/tex] we have
[tex]\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(q)\longleftrightarrow f(x)\in B_{\epsilon}(q)[/tex]

therefore we have to proof that [tex]\forall x'\in\mathbb{R}^{2}[/tex] we have
[tex]\forall \epsilon>0\exists\delta>0\mid \forall x\in B_{\delta}(x')\longleftrightarrow f(x)\in B_{\epsilon}(x')[/tex].

It's obvious that every element of [tex]\mathbb{R}^{2}[/tex] could be approximated by some element of [tex]\mathbb{Q}\times\mathbb{Q}[/tex] or sequence in this. But, how I can link this in an expression to get what I have to proof?
 
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  • #2
I'm not so sure about this but do we not know how the function maps irrational numbers, such as sqrt(2)?
 
  • #3
TimNguyen, in fact we know (they are mapped to themselves, as f is the identity map) but this is exactly what Elessar Telkont wants to show.

Indeed you got the idea right: any real number can be approximated by a sequence of rational numbers (and therefore, pairs of reals can be approximated by pairs of rationals).
What I would do is: Try to make this process of approximation precise (describe it in terms of epsilon-delta). Now assume what you want to prove is not true, then this should give a contradiction with the continuity (which you have also written out in epsilon-delta).

I will take a look and post it more precisely later on (first, you give it a try yourself)
 
  • #4
Hmm, it was much easier.

It is a familiar fact (or otherwise you should be able to easily prove it from the definition) that for continuous functions f, it holds that [tex]\lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n)[/tex] for a sequence [tex](x_n)_{n \in \mathbb{N}}[/tex].
So describe a real pair x as the (coordinate-wise) limit of a sequence [itex]x_n[/itex] of rational pairs, then
[tex]f(x) = f(\lim_{n \to \infty} x_n) \stackrel{*}{=} \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = x,[/tex]
where the identity marked with a star holds by the continuity of f -- QED.

[edit]For completeness, let me prove the claim about the limits (it's a nice exercise in epsilon-delta proofs, so you might want to try it yourself first):
Let [itex]\epsilon > 0[/itex]. Since f is continuous, there is some [itex]\delta[/itex] such that [itex]|| x - x_n || < \delta[/itex] implies that [itex]|| f(x_n) - f(x) || < \epsilon[/itex].
Now [itex]x_n[/itex] converging to x means that for this [itex]\delta[/itex] I can find an [itex]N[/itex] such that [itex]|| x_n - x || < \delta[/itex] as long as [itex]n > N[/itex].
So, through the [itex]\delta[/itex] from the definition of continuity, I have found an [itex]N[/itex] for my [itex]\epsilon[/itex] such that [itex] n > N[/itex] implies [itex]|| f(x_n) - f(x) || < \epsilon [/itex], in other words,
[tex] \lim_{n \to \infty} f(x_n) = f(x) = f( \lim_{n \to \infty} x )[/tex].
 
Last edited:
  • #5
CompuChip said:
TimNguyen, in fact we know (they are mapped to themselves, as f is the identity map) but this is exactly what Elessar Telkont wants to show.

Indeed you got the idea right: any real number can be approximated by a sequence of rational numbers (and therefore, pairs of reals can be approximated by pairs of rationals).
What I would do is: Try to make this process of approximation precise (describe it in terms of epsilon-delta). Now assume what you want to prove is not true, then this should give a contradiction with the continuity (which you have also written out in epsilon-delta).

I will take a look and post it more precisely later on (first, you give it a try yourself)

Sorry about that. My math is extremely rusty since I started graduate school in physics.
 

What is the definition of continuity for a function?

The definition of continuity for a function is that the function is continuous at a point if the limit of the function at that point is equal to the value of the function at that point. In other words, there are no abrupt changes or holes in the graph of the function at that point.

What is the difference between continuity and differentiability for a function?

Continuity refers to the smoothness of a function, while differentiability refers to the existence of the derivative of a function at a point. A function can be continuous without being differentiable, but a function cannot be differentiable without being continuous.

How can you determine if a function is continuous at a given point?

To determine if a function is continuous at a given point, you can use the definition of continuity to check if the limit of the function at that point is equal to the value of the function at that point. If they are equal, then the function is continuous at that point.

What are the three types of continuity for a function?

The three types of continuity for a function are pointwise continuity, uniform continuity, and local continuity. Pointwise continuity is when a function is continuous at each individual point in its domain. Uniform continuity is when a function is continuous over an entire interval. Local continuity is when a function is continuous in a neighborhood around a given point.

What are some common examples of discontinuous functions?

Some common examples of discontinuous functions include step functions, piecewise functions, and functions with removable or jump discontinuities. These types of functions have abrupt changes or breaks in their graphs that make them not continuous at certain points.

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