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f(x) = 8x - 5 is continuous on R

G-X

New member
Jul 11, 2020
21
Definitions: A function f with domain a subset D of \(\displaystyle R^{k}\) and range contained in \(\displaystyle R^{\ell}\), \(\displaystyle k = 1\) or \(\displaystyle 2\) and \(\displaystyle \ell = 1\) or \(\displaystyle 2\), is continuous at a point \(\displaystyle a \in D\) if for each \(\displaystyle \epsilon > 0\) there exists \(\displaystyle \delta > 0\) such that \(\displaystyle \forall x \in B_{\delta}(a) \cap D\), \(\displaystyle f(x) \in B \epsilon (f(a))\). The function is continuous on D if it is continuous at each point of D.

Let \(\displaystyle \epsilon > 0\) choose a \(\displaystyle \delta = \dfrac{\epsilon + 5}{8}\).

Let \(\displaystyle a \in R^{k}\) where \(\displaystyle k = 1\) or \(\displaystyle 2\). We can define an open ball \(\displaystyle B_{\delta}(a) = \){\(\displaystyle y \in R^{k}: d(y, a) < \delta\)}. Let \(\displaystyle x \in B_{\delta}(a) \cap R^{k}\).

Let \(\displaystyle b \in f(a)\). We can define another open ball \(\displaystyle B_{\epsilon}(b) = \){\(\displaystyle z \in R^{k}: d(z, b) < \epsilon\)}.

We need to prove that \(\displaystyle f(x) \in B\epsilon(b)\) which is equivalent to the statement \(\displaystyle B\epsilon(b) \cap f(x) = f(x)\). Since we want \(\displaystyle f(x)\) to be fully contained in the interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\). This is equivalent to the statement \(\displaystyle f(x) < \epsilon\). \(\displaystyle f(x) = 8x - 5 < 8\delta - 5 = 8(\dfrac{\epsilon + 5}{8}) - 5 = \epsilon + 5 - 5 = \epsilon\).

Thus we have proven that the function \(\displaystyle f(x) = 8x - 5\) is continuous in \(\displaystyle R^{k}\).

Additional Questions:
We know that x is in the range of \(\displaystyle (a - \delta, a + \delta)\). So x can be more than delta if not centered at 0. So a slight mistake in the proof?
It really should be that \(\displaystyle f(x) < \epsilon\) is \(\displaystyle -\epsilon< f(x) < \epsilon\), so the proof is incorrect?
The interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\) is only defined in \(\displaystyle R^{1}\). How do I define this concept in \(\displaystyle R^{2}\).
 
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G-X

New member
Jul 11, 2020
21
I can slightly change the proof, but then I might need to change \(\displaystyle \delta = \dfrac{\epsilon}{8}\), but then I don't think the delta works?

Definitions: A function f with domain a subset D of \(\displaystyle R^{k}\) and range contained in \(\displaystyle R^{\ell}\), \(\displaystyle k = 1\) or \(\displaystyle 2\) and \(\displaystyle \ell = 1\) or \(\displaystyle 2\), is continuous at a point \(\displaystyle a \in D\) if for each \(\displaystyle \epsilon > 0\) there exists \(\displaystyle \delta > 0\) such that \(\displaystyle \forall x \in B_{\delta}(a) \cap D\), \(\displaystyle f(x) \in B \epsilon (f(a))\). The function is continuous on D if it is continuous at each point of D.

Let \(\displaystyle \epsilon > 0\) choose a \(\displaystyle \delta = \dfrac{\epsilon + 5}{8}\).

Let \(\displaystyle a \in R^{k}\) where \(\displaystyle k = 1\) or \(\displaystyle 2\). We can define an open ball \(\displaystyle B_{\delta}(a) = \){\(\displaystyle y \in R^{k}: d(y, a) < \delta\)}. Let \(\displaystyle x \in B_{\delta}(a) \cap R^{k}\) where \(\displaystyle x < a + \delta\).

Let \(\displaystyle b \in f(a)\). We can define another open ball \(\displaystyle B_{\epsilon}(b) = \){\(\displaystyle z \in R^{k}: d(z, b) < \epsilon\)}.

We need to prove that \(\displaystyle f(x) \in B\epsilon(b)\) which is equivalent to the statement \(\displaystyle B\epsilon(b) \cap f(x) = f(x)\). Since we want \(\displaystyle f(x)\) to be fully contained in the interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\). This is equivalent to the statement \(\displaystyle -\epsilon < f(x) - f(a) < \epsilon\) or \(\displaystyle |f(x) - f(a)| < \epsilon\).

\(\displaystyle |f(x) - f(a)| = |8x - 5 - (8a - 5) |= |8(x - a)| < |8(a + \delta - a)| < 8\delta = 8(\dfrac{\epsilon + 5}{8}) = \epsilon + 5\).

Thus we have proven that the function \(\displaystyle f(x) = 8x - 5\) is continuous in \(\displaystyle R^{k}\).

Additional Questions:
The interval \(\displaystyle (f(a) - \epsilon, f(a) + \epsilon)\) is only defined in \(\displaystyle R^{1}\). How do I define this concept in \(\displaystyle R^{2}\). Maybe I already have?
 
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Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
A more important error is that you cannot talk about the interval (f(a)−ϵ, f(a)+ϵ) in \(\displaystyle R^2\). That notation is only for \(\displaystyle R^1\). Either use a generic notation that works for any n or do k= 1 and k= 2 separately.
 

G-X

New member
Jul 11, 2020
21
Does this proof work for R^2? I might need to change \(\displaystyle |x - a| < \delta\) to \(\displaystyle \sqrt{(x - a1)^{2} +(f(x) - a2)^{2}} < \delta\). If so, how do you create a proof that is universal? I think the proof works for k = 1 and k = 2 because the domain and range are broken up?

Definitions: A function f with domain a subset D of \(\displaystyle R^{k}\) and range contained in \(\displaystyle R^{\ell}\), \(\displaystyle k = 1\) or \(\displaystyle 2\) and \(\displaystyle \ell = 1\) or \(\displaystyle 2\), is continuous at a point \(\displaystyle a \in D\) if for each \(\displaystyle \epsilon > 0\) there exists \(\displaystyle \delta > 0\) such that \(\displaystyle \forall x \in B_{\delta}(a) \cap D\), \(\displaystyle f(x) \in B \epsilon (f(a))\). The function is continuous on D if it is continuous at each point of D.

Let \(\displaystyle \epsilon > 0\) choose a \(\displaystyle \delta = \dfrac{\epsilon}{8}\).

Let \(\displaystyle a \in R^{k}\) where \(\displaystyle k = 1\) or \(\displaystyle 2\). We can define an open ball \(\displaystyle B_{\delta}(a) = \){\(\displaystyle y \in R^{k}: d(y, a) < \delta\)}. Let \(\displaystyle x \in B_{\delta}(a) \cap R^{k}\) where \(\displaystyle |x - a| < \delta\).

Let \(\displaystyle b = f(a)\). We can define another open ball \(\displaystyle B_{\epsilon}(b) = \){\(\displaystyle z \in R^{k}: d(z, b) < \epsilon\)}.

We need to prove that \(\displaystyle f(x) \in B\epsilon(b)\) which is equivalent to the statement \(\displaystyle B\epsilon(b) \cap f(x) = f(x)\). Since we want \(\displaystyle f(x)\) to be fully contained in the interval \(\displaystyle B\epsilon(b)\). This is equivalent to the statement \(\displaystyle |f(x) - f(a)| < \epsilon\).

\(\displaystyle |f(x) - f(a)| = |8x - 5 - (8a - 5) |= |8(x - a)|\).

Case 1: \(\displaystyle \delta \ge 0: |8(x - a)| < |8(a + \delta - a)| < 8\delta = 8(\dfrac{\epsilon}{8}) = \epsilon\)
Case 2: \(\displaystyle \delta < 0: |8(x - a)| < |8(a - \delta - a)| < |-8\delta| = 8\delta = 8(\dfrac{\epsilon}{8}) = \epsilon\)

Thus we have proven that the function \(\displaystyle f(x) = 8x - 5\) is continuous in \(\displaystyle R^{k}\).
 
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G-X

New member
Jul 11, 2020
21
If x = (x1, x2). How does f(x) = 8x - 5 work? What is the result of f(x1, x2) = 8(x1, x2) - 5?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
In you very first post, you said " Thus we have proven that the function f(x)=8x−5f(x)=8x−5 is continuous in $R^k$". What does that even mean? What is "8x- 5" with x in $R^k$? I might guess that, with $x= (x_1, x_2, \cdot\cdot\cdot, x_k)$, $8x= (8x_1, 8x_2, \cdot\cdot\cdot, 8x_k)$ but what is "5"? You need to rethink what you are trying to prove!
 

G-X

New member
Jul 11, 2020
21
I don't think I have ever taken this type of math. From my understanding, with a domain that is in R^2, which again I have never heard of. That 5 would simply make f(x) undefined?

Unless, you allow 5 to become a magnitude or a vector... which in that case aren't you changing the nature of the equation?

Or do you ask yourself what is 5? Is it (5, 0) is it (0, 5)? Is it (5, 5)
 
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G-X

New member
Jul 11, 2020
21
If it is a set then does it become f(x) = {(x1, x2) - 5} - a variable that exists in R^1 only? and thus a goal would be to remove it? It would have a component in R^1 and a component in R^2? Thus there would be no algebraic operations that could be conducted on each other because they exist in separate coordinate spaces?
 

G-X

New member
Jul 11, 2020
21
From my understanding now, there can be four proofs depending upon your combination of the domain and range as it is just mapping the domain to the range if I am correct. So k = 1 or 2 and l = 1 or 2. Allowing:
k = 1, l = 1, k = 2, l = 1, k = 1, l = 2, k = 2, l = 2...

So how can R^1 and R^2 interact with each other? Set R^1 with a 0 component for R^2?

|x - a| < d only works for a k = 1 and a l = 1... or rather k = 1 because it is with regards to the domain...
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
I am getting more confused! What is it, exactly that you are trying to prove?



If it is to prove "f, from \(\displaystyle R^k\) to \(\displaystyle R^j\), is continuous" then what f is will depend on what k and j are,

"f(x)= 8x+5" can only be from \(\displaystyle R^1\) to \(\displaystyle R^1\).



If you want a function from \(\displaystyle R^1\) to \(\displaystyle R^2\) it must be something like \(\displaystyle f(x)= \begin{pmatrix}8x+ 5 \\ 6x- 9\end{pmatrix}\).

If you want a function from \(\displaystyle R^2\) to \(\displaystyle R^1\) it must be something like \(\displaystyle f\begin{pmatrix}x\\ y\end{pmatrix}= 3x- 7\).

If you want a function from \(\displaystyle R^2\) to \(\displaystyle R^2\) it must be something like \(\displaystyle f\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}3x- 5 \\ 4x+ 7 \end{pmatrix}\).
 

G-X

New member
Jul 11, 2020
21
How can f(x) = 8x + 5 only be from R^1 to R^1.

From my understanding, if you pick a domain that is in R^2. At least for this equation that forces the range to be R^2?

For instance, if f(x) = 8x and f(x) is a set then the only way for the range of f(x) to be in R^2 must be for the domain to be in R^2.

Let x = (x1, x2) in R^2 f((x1, x2)) = (8x1, 8x2).

Thus we have a function with a range in R^2.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
Then I will have to ask, again, if x is in $R^2$ then what do you mean by "8x+ 5"?
For example, if x= (2, 3), what is 8x+ 5?
 

G-X

New member
Jul 11, 2020
21
You mentioned that f(x) is just a set. This would mean that the below should be possible
Domain in R^2: x = (2, 3)
Range has just "complicated" entries: (16, 24) + 5

That is the range becomes components of R^2 and R^1 similar to imaginary/complex numbers. Existing in both R^2 and R^1


The problem is then you get these invalid entries and the only way to handle them would then to be to restrict the proof such that to k=1 and l = 1.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
You mentioned that f(x) is just a set. This would mean that the below should be possible
Domain: x = (2, 3)
Range: (16, 24) + 5

That is the range becomes components of R^2 and R^1 similar to imaginary/complex numbers.
Who is this in response to? I don't see where anyone "mentioned that f(x) is just a set".
Unfortunately, your notation is very confused and I am concerned that you might be confused about it. The domain of a function is a set. When you write "Domain: x= (2, 3)" I assumed you meant that the domain was the interval of real numbers from 2 to 3 (though "=" would be wrong there). But then you appear to be interpreting (2, 3) as the pair (2, 3) and then taking 8x to be (8*2, 8*3)= (16, 24). But then what do you mean by "(16, 24)+ 5"?

You mention "complex numbers". Do you intend (2, 3) to be 2+ 3i? That's an appropriate notation but "5" should be written as (5, 0) to be consistent. In that case 8(2, 3)+ (5, 0)= (16+ 5, 24)= (21, 24)- or, in more common complex numbers notation, 8(2+ 3i)+ 5= 15+ 5+ 24I= 21+ 24i.
 

G-X

New member
Jul 11, 2020
21
f(x) is operations on sets... does that mean that f(x) is a metric space where (x, f(x))? I have to read further into that, but...

In either case, this proof must be restricted to k = 1 and l = 1 then, correct?
 
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Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
f(x) is operations on sets... does that mean that f(x) is a metric space where (x, f(x))? I have to read further into that, but...

In either case, this proof must be restricted to k = 1 and l = 1 then, correct?
Where did you get "f(x) is operations on sets". And what do you mean by "where (x, f(x))?"? Where (x, f(x)) is what?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
485
Please answer at least some of the questions you have been asked! Mainly please answer "For x in $R^2$, what do you mean by "8x+ 5"? How are you defining "8x" for x in $R^2$? How are you adding "5" to that? Please help us understand what you are doing!