# Topology (Hocking and Young) two problems and their solutions

Reference: Chapter 1, Topology by Hocking and Young. Dover Publications, Inc. Available in Amazon India also.

Question 1:

Show that the mapping f used in the proof of the Theorem 1-18 is continuous.

Proof 1:

The mapping referred to f is the following:

Consider $f: E^{n+1} \rightarrow E^{n+1}$ carrying each point $(x_{1}, x_{2}, \ldots, x_{n+1}) \in E^{n+1}-0$ to $(\frac{x_{1}}{|x|^{2}}, \frac{x_{2}}{|x|^{2}}, \ldots, \frac{x_{n+1}}{|x|^{2}})$

We have to prove that this map f is continuous. Clearly, the $\epsilon - \delta$ definition of continuity will work here.

Consider any point $p \in E^{n+1}-0$ where $p = (\frac{x_{1}}{|x|^{2}}, \frac{x_{2}}{|x|^{2}}, \ldots, \frac{x_{n+1}}{|x|^{2}})$ where clearly $p \neq 0$

Let the n-tuple $x \in E^{n+1}$ and $x \neq 0$

Consider an $\epsilon$ ball, viz, $d(p,x) < \epsilon$ That is, we are considering a neighbourhood about the point x. Let the n-tuple x be given by $(y_{1}, y_{2}, \ldots, y_{n})$

Then, by definition of usual metric, $\sqrt{\frac{x_{1}-y_{1}}{|x|^{2}} + \frac{x_{2}-y_{2}}{|x|^{2}} + \ldots +\frac{x_{n+1}-y_{n+1}}{|x|^{2}}} < \epsilon$ where $|x|$ is the usual norm in $E^{n+1}$

So, by usual algebra, we get

$\sqrt{(x_{1}-y_{1})+(x_{2}-y_{2}) + \ldots + (x_{n+1}-y_{n+1})} < |x| \epsilon$. But this is an open ball of radius $|x|\epsilon$. So the given mapping f maps an open ball to open ball in given domain. Hence, f is open.

Question 2:

Given any closed interval $[a,b]$ in $E^{1}$ find a continuous mapping of $E^{1}$ into $[a,b]$ thereby proving that $[a,b]$ is connected.

Solution/Proof:

We prove the classic case $[a,b]=[0,1]$. Consider the set of all points x in the open unit interval $(0,1)$ is equivalent to the set of all points y on the whole real line. For example, the formula:

$y = \frac{1}{x}\arctan{x} + \frac{1}{2}$
establishes a 1-1 correspondence between these two sets. As this map is continuous, and it maps the real line, which is connected to the above interval, the given interval is also connected.

Cheers,

Nalin Pithwa

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