The set $\mathbb{R}^2-\{(0,0)\}$ with the usual topology is:
(A) Homeomorphic to the open unit disc in $\mathbb{R}^2$
(B) the cylinder $\{(x,y,z)\in \mathbb{R}^3/ x^2+y^2=1 \}$
(C) the paraboloid $\{(x,y,z)\in \mathbb{R}^3/ z=x^2+y^2 \} $
(D) the paraboloid $\{(x,y,z)\in \mathbb{R}^3/ z=xy \} $
I had tried which topological property is not hold under homeomorphism. I'm little bit confused in graph of (b),(c) and (d) because I have not any software related to 3d graphs. Please give me suggestion. Is this true $\mathbb{R}^2-\{(0,0)\}$ is homeomorphic to $S^2 = \{x\in \mathbb{R}^3/ \|x\|=1\}$ ? if it is not true then give some hint for option (a).Thanks in Advance.
The space $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $(a)$, $(c)$, or $(d)$, but it is homeomorphic to $(b)$.
To prove that $\mathbb{R}^2 \setminus \{(0,0)\}$ is homeomorphic to the cylinder $(b)$, define the following map. For the circle of radius $r$ around the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$, map this circle homeomorphically onto the circle of the cylinder at $z = \tan\left(\frac{\pi r}{r+1} - \frac{\pi}{2} \right)$. This covers all possible values of $z$, hence the whole cylinder.
Next, observe that $\mathbb{R}^2$ is homeomorphic to the topological spaces $(a)$, $(c)$, and $(d)$ via the maps:
$(a)$ Let $D$ denote the open unit disk. Then the map $D \to \mathbb{R}^2$ given by $x \mapsto x/(1-\|x\|)$ is a homeomorphism.
$(c)$ Let $P_1$ denote the paraboloid. Then the map $\mathbb{R}^2 \to P_1$ given by $(x,y) \mapsto (x,y,x^2+y^2)$ is a homeomorphism.
$(d)$ Let $P_2$ denote the paraboloid. Then the map $\mathbb{R}^2 \to P_1$ given by $(x,y) \mapsto (x,y,xy)$ is a homeomorphism.
Thus, to prove $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $(a)$, $(b)$, or $(c)$, it suffices to prove that $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $\mathbb{R}^2$.
To do this, we use the notion of homotopic curves. In a topological space $X$, two closed curves (continuous maps $\gamma : [0,1] \to X$ with $\gamma(0) = \gamma(1)$) are homotopic if they can be deformed continuously from one to the other. For instance, in $\mathbb{R}^2$, the unit circle and the circle of radius 2 (centered at the origin) are homotopic, by translating each point on the unit circle to its corresponding point on the circle of radius 2, along the line connecting them. Likewise, the unit circle is homotopic to the constant curve at the origin.
More opaquely, a homotopy between closed curves $\gamma_1$ and $\gamma_2$ is a continuous map $H:[0,1]\times[0,1] \to X$ such that $H(0,t) = \gamma_1(t)$ and $H(1,t) = \gamma_2(t)$ for all $t\in [0,1]$.
Since homotopies are continuous deformations, if $f: X \to Y$ is a homeomorphism and $\gamma_1$, $\gamma_2$ are homotopic in $X$, then the closed curves $f\circ\gamma_1$ and $f\circ\gamma_2$ are homotopic in $Y$.
Let us use this fact. The intuition is that any closed curve in $\mathbb{R}^2$ is homotopic to a point, whereas this is not true for closed curves enclosing the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$. We will prove this latter fact now.
Lemma: A closed curve enclosing the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homotopic to a point.
Proof: Suppose $\gamma_1$ is a closed curve in $\mathbb{R}^2 \setminus \{(0,0)\}$ enclosing the origin, and $\gamma_2$ is the constant curve at some point $c\in\mathbb{R}^2$. For the sake of contradiction, assume $H(s,t)$ is a homotopy between $\gamma_1$ and $\gamma_2$. Let $I$ be the subset of $[0,1]$ such that for every $x\in I$, the closed curve $H(x,t)$ encloses the origin. Since $[0,1]$ is compact, we see that the distance $|H(x,t) - H(x+\delta, t)|$ reaches a maximum as $t$ ranges over $[0,1]$, and this maximum can be made as small as desired by changing $\delta$. Hence, we can choose $\delta$ so that if $x\in I$ then $x+\delta \in I$. It follows that $I$ is an open subset of $[0,1]$. Now, take a convergent sequence $\{x_n\} \to x$ in $I$. If the distance $|H(x_n, t)|$ is bounded below as $t$ ranges over $[0,1]$, then $H(x,t)$ must also enclose the origin. If this distance gets arbitrarily small: let $t_n \in [0,1]$ be such that $|H(x_n, t_n)|$ is smallest. Since $[0,1]$ is compact, there is a convergent subsequence $t_{n_k} \to t$. Then $|H(x,t)| = 0$, which is a contradiction. Thus, $I$ is a closed subset of $[0,1]$. As $[0,1]$ is connected, $I$ must be all of $[0,1]$, and thus the homotopy cannot exist.
At last, we can prove $\mathbb{R}^2$ is not homeomorphic to $\mathbb{R}^2 \setminus \{(0,0)\}$. Suppose $f: \mathbb{R}^2 \setminus \{(0,0)\} \to \mathbb{R}^2$ is a homeomorphism. Choose a closed curve $\gamma$ enclosing the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$. Let $H$ be a homotopy between $f\circ \gamma$ and a point in $\mathbb{R}^2$. Then $f^{-1} \circ H$ is a homotopy between $\gamma$ and a point in $\mathbb{R}^2 \setminus \{(0,0)\}$. By the lemma, this is impossible.
The ideas here regarding homotopy of curves form the basis for algebraic topology, and the theory of the fundamental group. The fundamental group of a topological space $X$ is the set of equivalence classes of closed curves, under the equivalence relation of homotopy. What we proved here is essentially that the fundamental groups of $\mathbb{R}^2$ and $\mathbb{R}^2 \setminus \{(0,0)\}$ are nonisomorphic.
As a closing remark, the fundamental group of $\mathbb{R}^2$ is 0 because every closed curve is homotopic to a point. What is the fundamental group of $\mathbb{R}^2 \setminus \{(0,0)\}$? Can you prove your guess?