$[0,1)$ as a subspace of the Euclidean metric space? Consider the Euclidian metric space $(\Bbb R,d)$ where $d(x,y)=|x-y|$ is the usual metcic on $\Bbb R$. The set $[0,1)$ is not closed in $(\Bbb R,d)$ but considered as a subspace it is closed by definition so that in $([0,1),d)$, $[0,1)$ is closed.
But now comes the question, $1\notin [0,1)$ but is an accumulation point since $x_n = 1-{1\over n}$ is a sequence in $[0,1)$ that converges to $1$ even in the subspace metric. So how can $[0,1)$ be considered closed even in the subspace topology when we clearly have the well known statement that, a set is closed iff it contains all its accumulation points? Thanks.
 A: The reason why $x_{n}$ does not converge to $1$ in the subspace metric is that $1$ is not in the subspace. So $1$ is not an accumulation point of $[0,1)$ in its own subspace topology, simply because $1\notin [0,1)$. It is true however that $(x_{n})$ is a Cauchy sequence in the subspace, but this sequence has no limit. 
A: Recall the definition of an accumulation point. If $(X,d)$ is a metric space and $A \subseteq X$, then $x \in X$ is an accumulation point of $A$ if there exists a sequence $(x_n)$ in $A$ converging to $x$. The important part here is that every accumulation point lies in $X$ by definition.
So if you take $X = A = [0,1)$ then $1$ is simply no accumulation point of $A$ because it is not contained in $X$.
Also note that the sequence $x_n = 1 - \frac{1}{n}$ does not converge in $X = [0,1)$, because there is no $x \in X$ such that $d(x_n,x) \to 0$.
A: $1$ is not an accumulation point, given that do not belong to the space. The space is closed (a whole topological space is always closed) but not complete, given that the give sequence is a Cauchy sequence but does not converge.
