Show that the function $g: (0,1) \rightarrow \mathbb{R}$ given below is continuous on irrationals and discontinuous on rationals:

$g(x) = \frac{1}{q} \; if \; x \in \mathbb{Q} \; \cap \; (0,1)$and $x = \frac{p}{q}$ in reduced

form = 0 otherwise.

We say that a set is a $G_{\delta}$ set if it is countable intersection of open sets.

Show that a bijective continuous map from a compact metric space into a metric space sends closed sets to closed sets, and hence it is a homeomorphism.

Define $f: \mathbb{R} \rightarrow [0,1]$ by $f(x) =x$ if $|x| \le 1,$ and $f(x)= \frac{1}{|x|}$ if $|x| \ge1$. Then f is continuous on $\mathbb{R}$. Note that $f$ is onto but not one-one.

 Show that the interval $(a,b) \subseteq \mathbb{R}$ is homeomorphic to any other interval $(c,d) \subseteq \mathbb{R}$.

Show that $e^{-x}$ is a homeomorphism from $(0, \infty)$ onto $(0,1)$.

Which of spaces $X$ and $Y$ are homeomorphic:

(1) $X= \mathbb{R}$ and $Y=[0,1)$

(2) $X=\mathbb{R}$ and $Y=[0,1]$

(3) $X=[1, \infty)$ and $Y=(0,1]$

(4) $X=\mathbb{R}$ and $Y=(0,1)$

(5) $X=\mathbb{Q}$ and $Y=\mathbb{Q}$

(6) $X=\{(x,y,z) \in \mathbb{R^3} : x^2+y^2 =z ,r<z<R \}$ and $Y =A(0,r,R)$

(7) $X= \{x \in \mathbb{R^{n+1}}:x_{n+1}=0 \}$ and $Y= \mathbb{R^n}$

Verify that $\phi(t)=e^{it}$ is a homeomorphism between $(0,2\pi)$ and $\mathbb{T} \backslash \{1\}$, where $\mathbb{T}$ denotes the unit circle in the complex plane. Conclude that unit circle minus a point is homeomorphic to the real line.

Show that $X=A(0,r,R)$ (open annulus) and $Y= \mathbb{T}$(unit circle) are not homeomorphic

Let $X=S^n \backslash \{(0,\ldots ,1)\} \subseteq \mathbb{R^{n+1}}$ and $Y= \mathbb{R^n}$ , and define

$f:X \rightarrow Y$ and $g:Y \rightarrow X$ by

$f(x_1, \ldots ,x_{n+1}) =\large( \frac{x_1}{1-x_{n+1}}, \ldots , \frac{x_n}{1-x_{n+1}}),\\g(x_1, \ldots ,x_n)=\large(\frac{2x_1}{{||x||_2}^2+1} ,\ldots ,\frac{2x_n}{{||x||_2}^2+1} , \frac{{||x||_2}^2-1}{{||x||_2}^2+1} )$

Verify that $f$ and $g$ are homeomorphisms such that $f^{-1}=g$.

Show that the graph $G$ of a continuous function $f:X \rightarrow Y$ (with the subspace topology inherited from $X \times Y$ ) is homeomorphic to $X$.

Consider the continuous function $f: \mathbb{R} \backslash \{0\} \rightarrow \mathbb{R}$ by $f(x)=1/x$. The graph of f is the hyperbola $xy=1$. One may conclude from the last exercise that $\mathbb{R} \backslash \{0\}$ is homeomorphic to $xy=1$.