- Thread starter
- #1

- Mar 10, 2012

- 835

Please check whether the following statement is true:

Lemma:Let $X$ be any Hausdorff space. Let $U$ be open in $X$. Let $y \in U$. Then $U \setminus \{ y \}$ is open in $X$.

Proof: Write $V=U \setminus \{ a \}$. Let $x \in V$. Then $\exists$ open sets $A$ and $B$ of $X$ such that $x \in A , y \in B$ and $A \cap B = \emptyset$. Then $A \cap U$ is an open in $X$ which is a subset of $V$ and contains $x$. This can be done for all $x \in V$ thus $V$ is open.

Assuming the above is fine, I need to prove this:

Let $p$ be a point in $S^n$.

Let $h_1 : B^n - S^{n-1} \rightarrow \mathbb{R}^n$ and $h_2 : \mathbb{R}^n \rightarrow S^n - \{ p \}$ be homeomorphisms.

Define $f : B^n \rightarrow S^n$ as:

$f(x)=h_2 h_1 (x) \, \, \text{ for } x \in B^n - S^{n-1}, f(x)=p \, \, \text{ for } x \in S^{n-1}$.

Show that $f$ is continuous.

Attempt:

Let $U$ be open in $S^n$. If $p \notin U$ then its easy to show that $f^{-1}(U)$ is open in $B^n$.

Here's the problem case. Assume $p \in U$. Clearly $S^n$ is Hausdorff, so by the statement proved above $U- \{ p \}$ is open in $S^n$.

So $f^{-1}(U)=f^{-1}(U-\{ p \}) \cup f^{-1}(p)= \text{ (an open set in B^n)} \cup S^{n-1}$.

If $S^{n-1}$ were open in $B^n$ I'd be done, but it ain't. What do I do?

NOTE: $B^n$ is the unit ball in $\mathbb{R}^n$ and $S^{n-1}$ is its boundary.

Lemma:Let $X$ be any Hausdorff space. Let $U$ be open in $X$. Let $y \in U$. Then $U \setminus \{ y \}$ is open in $X$.

Proof: Write $V=U \setminus \{ a \}$. Let $x \in V$. Then $\exists$ open sets $A$ and $B$ of $X$ such that $x \in A , y \in B$ and $A \cap B = \emptyset$. Then $A \cap U$ is an open in $X$ which is a subset of $V$ and contains $x$. This can be done for all $x \in V$ thus $V$ is open.

Assuming the above is fine, I need to prove this:

Let $p$ be a point in $S^n$.

Let $h_1 : B^n - S^{n-1} \rightarrow \mathbb{R}^n$ and $h_2 : \mathbb{R}^n \rightarrow S^n - \{ p \}$ be homeomorphisms.

Define $f : B^n \rightarrow S^n$ as:

$f(x)=h_2 h_1 (x) \, \, \text{ for } x \in B^n - S^{n-1}, f(x)=p \, \, \text{ for } x \in S^{n-1}$.

Show that $f$ is continuous.

Attempt:

Let $U$ be open in $S^n$. If $p \notin U$ then its easy to show that $f^{-1}(U)$ is open in $B^n$.

Here's the problem case. Assume $p \in U$. Clearly $S^n$ is Hausdorff, so by the statement proved above $U- \{ p \}$ is open in $S^n$.

So $f^{-1}(U)=f^{-1}(U-\{ p \}) \cup f^{-1}(p)= \text{ (an open set in B^n)} \cup S^{n-1}$.

If $S^{n-1}$ were open in $B^n$ I'd be done, but it ain't. What do I do?

NOTE: $B^n$ is the unit ball in $\mathbb{R}^n$ and $S^{n-1}$ is its boundary.

Last edited: