Misunderstanding with the local definition of continuity I know that for any two top spaces $(X,\tau_X),(Y,\tau_Y)$ a function $f:X\to Y$ is said to be continuous on $X$ if $f^{-1}(V)\in\tau_X~\forall~V\in\tau_Y.$ Following such a definition the continuity of a function $f:A(\subset X)\to Y$ on $D\subset A$ should be as follows: $f$ is continuous on $D$ if $f|_{D}^{-1}(V)\in\tau_{X_D}~\forall$ $V\in Y$ where $\tau_{X_D}$ is the topology on $D$ relative to the topology $\tau_X$ on $X.$
Well! Let's come to the local case. Let $x\in X.$ Shouldn't the continuity of $f$ at $x$ be as follows: $f$ is continuous at $x$ if $f|_{\{x\}}:\{x\}\to Y$ is continuous on $\{x\}?$ 
But with respect to the relative topology of $X$ the topology of $\{x\}$ is $\{\{x\},\emptyset\}$ which in term implies then irrespective of what $f$ we choose it's locally continuous whereever defined!
So where did I go wrong?
 A: Continuity of $f$ on the subspace $\{x\}$ would be precisely what you describe. For continuity at a point, we'd like the general definition to agree with the notion of continuity that's more familiar (from metric spaces, and from real analysis in particular)--namely, that $\lim_{t\to x}f(t)=f(x)$.
Of course, in a general topology there may be no notion of limit in some of the usual senses (sequences may converge everywhere, for example), but in general, what we want to be able to say is: in order to make $f(t)$ as "close" to $f(x)$ as we like--within whatever neighborhood of $f(x)$ that we choose--we need only select $t\in X$ that is "close" enough to $X$--within some neighborhood of $x$. So, $f$ is continuous at $x$ if for any $V\in\tau_Y$ with $f(x)\in V$, there is some $U\in \tau_X$ with $x\in U$ and $f(U)\subseteq V$. Equivalently, for any $V\in\tau_Y$ with $f(x)\in V$, we have that $f^{-1}(V)\in\tau_X$.
As a side note, if $x$ is an isolated point of $X$--that is, if $\{x\}\in\tau_X$, then continuity on $\{x\}$ and continuity at $x$ coincide.
A: What you say is all completely correct, the "problem" arises depending on how you view continuity at a point. In fact, a continuous function $f:X\rightarrow Y$ is continuous at $x \in X$ if for any open set $V \in \tau_Y$ that contains $f(x)$, $f^{-1}(V)$ is open in $X$. 
Your definition using the subspace topology is valid as a definition of continuity for a function on the single-point space ${x}$, but, as you say, causes some problems if we want to think of $x$ as embedded in a larger apce. This is essentially because when we use the subspace topology, we "lose information" about the rest of the space. Consider the real-valued function on $\mathbb{R}$ given by $f(x) = 0$ if $x=0$ and $f(x) = 1$ otherwise. We can see that $f^{-1}(B_{0}(\frac12)) = 0$ which will be open in the subspace topology of ${0} \subset X$ but not in $X$ itself.
A: $f$ is indeed continuous on the set $\{x\}$, using the definition of continuity plus the definition of subspace topology. This is quite trivial. And true for any $f$ at all.
But we want $f$ continuous at $x$ to have a meaning that is consistent with the statement, that $f$ is continuous on $X$ iff $f$ is continuous at $x$ for all $x$ in $X$.
This is achieved by defining $f$ to be continuous at $x$ iff for every open neighbourhood $V$ of $f(x)$ there is an open neighbourhood $U$ of $x$ such that $f[U] \subset V$, which niceley translates to the usual $\epsilon$-$\delta$ definition for metric spaces.
