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I have a function that goes as following

$$f:\mathbb{R^6}\to\mathbb{R}^6\\(x_1,x_2,x_3,y_1,y_2,y_3)\mapsto\frac{x^2_2+y^2_2-x_1x_2+x_1x_3+y_1y_2+y_1y_3-x_2x_3-y_2y_3}{(x_3-x_2)^2+(y_3-y_2)^2}$$

since we're working with reals, it suffices to show that if $f=\frac{h}{g}$, if $h$ is and $g$ is continuous then $f$ is also continuous. But I already struggle to show that any of them are continuous. Let $p=(u_1,u_2,u_3,v_1,v_2,v_3)\in \mathbb{R}^6$ then $(x_1,x_2,x_3,y_1,y_2,y_3)\in h^{-1}(u_1,u_2,u_3,v_1,v_2,v_3)$ I will also denote $u_0=(u_1,u_2,u_3)$ and $v_0=(v_1,v_2,v_3)$ and we have $p_u<h(u_0,v_0)<p_v$ meaning we can find a $\varepsilon >0$ such that

$$p_u+\varepsilon <h(u_0,v_0)<p_v-\varepsilon$$

let $z=(u_0,v_0)$ then we have $B\left(z,\varepsilon\right)\subset h^{-1}(u_0,v_0)$, now I would introduce $(u_0',v_0')\in B(z,\varepsilon)$ then we would have $|u_0-u_0'|<\varepsilon$ same for $|v_0-v_0'|<\varepsilon$. But somehow this does not feel right at all. If I can show that $B(z,\varepsilon)\subset h^{-1}(u_0,v_0)\implies h^{-1}(u_0,v_0)$ is open and hence $h$ is cotninuous..... I want to prove continuity using open sets, but I don't see how with this given function.

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  • $\begingroup$ The function is going to $\Bbb R$ not $\Bbb R^6$. Not that it matters for the argument. $\endgroup$ Oct 20 at 16:15
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This was meant to be a comment but those do not accept lists.

  1. Projections are continuous, i.e. functions of the form $\mathbf{x} \to x_i.$
  2. Sums, products of continuous is continuous. Quotients of continuous are continuous wherever they are defined. (That is, wherever the denominator/bottom is not zero.)
  3. Thus, polynomials and rational functions are continuous.
  4. You said you want to prove continuity by open sets, I highly doubt this is a homework exercise (if it is, it is a unpedagogical one). If it is a self-study problem, then you aren't "getting it." The point of proving theorems or partial results is to break a big problem into smaller more handleable pieces (every mathematician and scientist works this way in one form or another).

Notice that the function you are defining is not defined on the set where $x_2 = x_3$ and $y_2 = y_3.$ (Its domain is not $\mathbf{R}^6.$) It is continuous elsewhere for the reasons given above.

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  • $\begingroup$ This was actually in the space $\mathbb{C}^3$ which is isomoprhic to $\mathbb{R}^6$. I failed to mention that in the post. Each point $(x,y,z)\in\mathbb{C}^3$ are distinct $(x\neq y\neq z \neq x)$ in $\mathbb{C}^3$ I have $f(x,y,z)=(z-x)/(y-x)$, idk if this translates that in reals that each $x_1,x_2...$ are distinct too. But. it felt natural to go to $\mathbb{R}$ instead of $\mathbb{C}$ when proving continuity. $\endgroup$ Oct 20 at 15:31
  • $\begingroup$ this is a homework problem, where I have to show that my function $f$ is a homeomorphism. I've already shown bijection and plan to show that $f$ is an open mapping to show that the preimage of $f$ exists and is also continuous. $\endgroup$ Oct 20 at 15:34
  • $\begingroup$ also this is not the entire function, I've split it up into 3 components i.e $f=(f_1,f_2,f_3)$, $f_1,f_2$ i've sucsessfully shown that they are continuous, but its this $f_3=(z-x)/(y-x)=((x_1+iy_1)-(x_2+iy_2)/((x_3+iy_3)-(x_2+iy_2))$ and since I'm working on $\mathbb{R}^6$ . I'm only interested in $\Re{(f_3)}$ $\endgroup$ Oct 20 at 15:37
  • $\begingroup$ Even if $(x_i, y_i)$ was a complex number, your function would still not be defined everywhere. Indeed, if $f(x,y,z) = (z-x)/(y-x)$ for complex numbers, then the function is defined on $\mathbf{C} \setminus \{x = y\}.$ And it is a rational function elsewhere, thus continuous. $\endgroup$
    – Will M.
    Oct 20 at 15:40
  • $\begingroup$ How do you justify that it's rational elsewhere? $\endgroup$ Oct 20 at 15:46

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