Let recip={(x,y) : x is a number different than 0 and y=1/x}
Show that recip is continuous on {x : x is a number different than 0}
I would not be surprised if the following is an incorrect interpretation, but I think you're asking to prove that $\displaystyle \varphi:\mathbb{R}-\{0\}\to\mathbb{R}\times\mathbb{R}:x\mapsto\left(x ,\frac{1}{x}\right)$ is continuous.
But, depending on your background knowledge this should be obvious since in any topological space if $\displaystyle \varphi,\psi:X\to Y$ are continuous maps then so is $\displaystyle \varphi\oplus\psi:X\to Y\times Y:x\mapsto (\varphi(x),\psi(x))$ this follows easily since $\displaystyle (\varphi\oplus\psi)^{-1}(U\times V)=\varphi^{-1}(U)\cap\psi^{-1}(V)$.
So, since $\displaystyle \varphi:\mathbb{R}-\{0\}:x\mapsto\frac{1}{x}$ and the inclusion map $\displaystyle \iota:\mathbb{R}-\{0\}\hookrightarrow\mathbb{R}$ are continuous then so is $\displaystyle \iota\oplus\varphi:\mathbb{R}-\{0\}\to\mathbb{R}\times\mathbb{R}:x\mapsto\left(x ,\frac{1}{x}\right)$
Another way is to notice that the two coordinate functions are continuous.
Well, for this problem, we are only working with limited knowledge and other theorems we've proven. Pretty much all we have to work with is assuming recip is a function and the definition of continuous that we are given:
Suppose X is a subset of the reals, f is a function from X into the reals, and p is an element of X. The statement that f is continuous at p means that if V is segment and f(p) is an element of V, then there is a segment, call such segment, call such a segment U, so that
i) p is an element of U,
and
ii)if q is an element of U intersect X, then f(q) is an element of V
The statement that f is continuous on X means that if p is and element of X, then f is continuous at p.
Sorry for lack of symbols, I'm new.
The way I understand it is have some segment (a,b) = V so that a<f(p)<b or for this problem a<1/p<b
Then show there exists a segment U such that p must be an element of U. Usually its just through algebra with the inequality to solve for p and make sure there can be a logical segment that p is an element of and is a subset of the domain.
I don't know if that helps, but I'm not fan of the definition either.
It sounds as if the professor was a student of one of Herbert Wall’s students.
An open finite non-degenerate interval is called a segment.
If $\displaystyle f(p)\in (c,d)$ then there is an $\displaystyle U_{\delta}(p)=\left(p-\delta,p+\delta\right)$ so that $\displaystyle f( U_{\delta}(p))\subset (c,d) $.
Actually this question just asks that we prove that if $\displaystyle p\not= 0$ then $\displaystyle \frac{1}{x}$ is continuous at $\displaystyle p$.
Here's what I thought would've worked, but not for all cases as I found out. Maybe this will help show what I'm talking about.
Pf: 2 Cases p>0 or p<0 Let f(x)=recip(x)
Let p>0 and p is an element of the reals. Let X be the set of postive real numbers such that p is an element of X.
Let V be the segment _(a,b)_ [i found out this only worked for me if a and b were positive] and let U be the segement _(1/b, 1/a)_
Then a<f(p)<b
a<1/p<b
1/p>p>1/b and thus p is an element of U
Let q be an element of U intersect X which impies
1/b<q<1/a
b>1/q>a
b>f(q)>a and thus f(q) is an element of V
Therefore, recip is continuous on X. [ ]
Similar proof for p<0
Unfortunately I thought this worked, but I think about V being the segment _(a,b)_ if a was negative and b was positive.
I'm still just stuck on this one.
P.S. Is there a chart for all the [tex] commands?