$C(X) := X \times [0,1]/ \sim $, when $(x_1,0) \sim (x_2,0)$ is connected. Let $X$ a topological space. 
We define $C(X) := X \times [0,1]/ \sim $, when $(x_1,0) \sim (x_2,0)$ for every $x_1, x_2 \in X$. Let $X$ any topological space. Prove: 
(a) $C(X)$ is connected, for any topological space $X$.
(b) $C(X)$ is Hausdorff iff $X$ is Hausdorff.
(c) $C(X)$ is compacted iff $X$ is compacted.
 A: Let $p:X\times [0,1]\rightarrow C(X)$ the quotient map.
a) Let $f:C(X)\rightarrow \{0,1\}$ be a continuous map. Write $g=f\circ p$.
$g$ is continue. For every $x\in X$, $g_{\mid x\times [0,1]}$ is constant since $x\times [0,1]$ is connected. This implies that $g(x,t)=g(x,0), t\in [0,1]$. If $y$ is another element of $X$, $f(p(y,t)))=g(y,t)=g(y,0)=f(p(y,0))=f(p(x,0))=g(x,0)=g(x,t)=f(p(x,t))$. This implies that $g$ is constant and $f$ are constant since $p$ is surjective.
b)
Suppose that $X$ is separated, $X\times [0,1]$ is separated. Since the restriction of $p_{\mid X\times (0,1])}:X\times (0,1]\rightarrow p(X\times (0,1])$ is an homeomorphism, we deduce that $p(X\times (0,1])$ is separated. Let $t\neq 0$, $x,y\in X, x\neq y$ there exist open subsets $U$, $V$ of $X$ such that $x\in U, y\in V, U\cap V=\phi$. There exists open subsets $0\in W_1, t\in W_2$ with $W_1\cap W_2=\phi$. $p(U\times W_1)\cap p(V\times W_2)=\phi$. We deduce that $C(X)$ is separated. 
Suppose that $C(X)$ is separated, consider $t\neq 0$ and $i_t:X\rightarrow X\times [0,1]$ defined by $i_t(x)=(x,t)$. Let $x,y\in X, x\neq y$.  There exists $U,V$ open subsets of $C(X)$ such that $p(x,t)\in U, p(y,t)\in V$ and $U\cap V=\phi$. Write $U'=(p\circ i_t)^{-1}(U), V'=(p\circ i_t)^{-1}(V)$, $x\in U, y\in V$ and $U'\cap V'=\phi$ since $p\circ i_t$ is injective. We deduce that $X$ is separated.
c)
Suppose that $X$ is compact, $X\times [0,1]$ is compact since it is the product of compact. We deduce that $C(X)=p(X\times [0,1])$ is compact since the image of a compact by a continuous map is compact.
Suppose that $C(X)$ is compact. Let $t>0$ $p(X\times t)$ is closed, since the inverse image of its complementary subset by $p$ is $X\times ([0,t)\bigcup (t,1])$. We deduce that $p(X\times t)$ is compact. Since the restriction of $p$ to $X\times t$ in an homeomorphism onto $p(X\times t)$, we deduce that $X\times t$ and henceforth $X$ is compact.
