Open/closed sets in $[ 0, 1 ]$ Let $X$ be the unit interval $[ 0, 1 ]$ with the usual topology. What is the nature of $[ 0, 1 )$ and $\{ 1 \}$ in $X$?
$\mathrm{Cl} ( [ 0, 1 ) ) = [0,1] \Rightarrow [ 0, 1 )$ is not closed.
$\mathrm{Int} ( [ 0, 1 ) ) = -\mathrm{Cl} ( - [ 0, 1 ) ) = -\mathrm{Cl} ( \{ 1 \} ) = -\{ 1 \} = [ 0, 1 ) \Rightarrow [ 0, 1 )$ is open?

$\mathrm{Cl} ( \{ 1 \} ) = \{ 1 \} \Rightarrow \{ 1 \}$ closed;
$\mathrm{Int} ( \{ 1 \} ) = - \mathrm{Cl} (- \{ 1 \} ) = - \mathrm{Cl} ( [ 0, 1 ) ) = - [ 0, 1 ] = \phi \Rightarrow \{ 1 \}$ is not open?
If my reasoning is right, $C= \{ [ 0, 1 ), \{ 1 \} \}$ is neither a closed, nor an open cover of $X$. But I have seen enough examples of homotopy where
$$
  F(x,t) =
  \begin{cases}
    x    &\textrm{if } t = 1\\
    f(x) &\textrm{if } 0 \leq t < 1,
  \end{cases}
$$
where $f(x)$ is some continuous function. $F(x,t)$ is claimed continuous. But a piecewise function is continuous iff $C$ is an open cover/finite closed over.
So, my reasoning above is obviously wrong, but I can't figure where. Any help would be great, thanks!
 A: The open sets in $I=[0,1]$ are obtained intesecting $I$ with open sets of $\Bbb R$. Any open set $A$ containing $1$ will contain some open interval $(1-\epsilon,1+\epsilon)$, so it is impossible that $A\cap I=\{1\}$.
On the other hand $\{1\}$ is closed in $I$, because it is closed in $\Bbb R$.
About $J=[0,1)$, can you find an open set $B\subset\Bbb R$ such that $B\cap I=J$?
A: Your reasoning is correct: $[0,1)$ is open in $[0,1]$ (e.g. because its complement in $[0,1]$ , namely $\{1\}$ is closed (as all singletons in a metric space)), but not closed, etc.
So indeed the continuity of a description as your example $$F(x,t) =
  \begin{cases}
    x    &\textrm{if } t = 1\\
    f(x) &\textrm{if } 0 \leq t < 1,
  \end{cases}$$
does not follow from the standard theorem that a function defined via partial function on a (locally) finite closed cover, or any open cover is continuous iff its parts are continuous.
The openness or closedness of the cover is a sufficient condition, not a necessary one.
In the example one would have to show continuity by another way. To have continuity it is necessary that $f(x) = F(x, t_n) \rightarrow F(x, 1) = x$ when $t_n \rightarrow 1$, so $f(x) = x$ for all $x$ is necessary for continuity... 
