Remark 3

Let $X$ be a topological space, and for each $x \in X$ let $\mathfrak{B}(x)$ be the set of all neighbourhoods of $x$. Let $x_0$ be a point of $X$; for each $x \in X$, define a set $\mathfrak{B}_0(x)$ of subsets of $X$ as follows: $\mathfrak{B}_0(x_0) = \mathfrak{B}(x_0)$, and if $x \neq x_0$ then $\mathfrak{B}_0(x)$ is to be the set of all subsets of $X$ which contain $x$. It is immediately verified (§ 1, no. 2, Proposition 2) that the sets $\mathfrak{B}_0(x)$ are sets of neighbourhoods of points of $X$ for a topology on $X$; let $X_0$ denote the topological space thus obtained, and let $j : X_0 \to X$ denote the identity map, which is continuous but not in general bicontinuous. A mapping $f$ of $X$ into a topological space $X'$ is continuous at the point $x_0$ if and only if the composition $X_0 \xrightarrow{j} X \xrightarrow{f} X'$ is continuous on $X_0$; this follows immediately from the definitions.