# Show that a mapping is continuous

1. Sep 29, 2012

1. The problem statement, all variables and given/known data
Show that the mapping f carrying each point $(x_{1},x_{2},...,x_{n+1})$ of $E^{n+1}-0$ onto the point $(\frac{x_{1}}{|x|^{2}},...,\frac{x_{n+1}}{|x|^{2}})$ is continuous.

2. Continuity theorems I am given.
A transformation f:S->T is continuous provided that if p is a limit point of a subset X of S then f(p) is a limit point or a point of f(X).

Let f:S->T be a transformation of space S into space T. A necessary and sufficient condition that f be continuous is that if O is an open subset of T, then its inverse image $f^{-1}(O)$ is open in S.

A necessary an sufficient condition that the transformation f:S->T be continuous is that if x is a point of S, and V is an open subset of T containing f(x) then there is an open set U in S containing x and such that f(U) lies in V.

3. The attempt at a solution
I was thinking to prove this I would have to find an open set in the range of this function and show that its inverse image is also open in the domain but I am not sure how I would go about doing that.

Any help would be appreciated.

Last edited: Sep 29, 2012
2. Sep 29, 2012

### voko

What is the definition of continuity you work with?

3. Sep 29, 2012

I have three. They are

A transformation f:S->T is continuous provided that if p is a limit point of a subset X of S then f(p) is a limit point or a point of f(X).

Let f:S->T be a transformation of space S into space T. A necessary and sufficient condition that f be continuous is that if O is an open subset of T, then its inverse image f^(-1)(O) is open in S.

A necessary an sufficient condition that the transformation f:S->T be continuous is that if x is a point of S, and V is an open subset of T containing f(x) then there is an open set U in S containing x and such that f(U) lies in V.

I updated the original post to reflect this.

4. Sep 29, 2012

### voko

I would use the first one. Given a sequence $\mathbf{x}^{(n)} \rightarrow \mathbf{p}$, prove that $\frac {\mathbf{x}^{(n)}} {(x^{(n)})^2} \rightarrow \frac {\mathbf{p}} {p^2}$.

5. Sep 29, 2012

I don't really understand. Can you elaborate a little bit? I don't know how I would show that a point $x=(x_{1},x_{2},...,x_{n+1})$ is a limit point.
You don't have to prove $\mathbf{x}^{(n)} \rightarrow \mathbf{p}$, this is an assumption. Given that assumption, prove $\frac {\mathbf{x}^{(n)}} {(x^{(n)})^2} \rightarrow \frac {\mathbf{p}} {p^2}$.