Continuity was defined for a map as a whole. It can also be asked at a single point: whatever tolerance is demanded around the value, there is room around the point that the map sends inside it. A map is continuous exactly when it is continuous at every one of its points, and this local form is the one that recovers the epsilon and delta of the real line.

A dial is reliable at a setting if, for every accuracy you demand at the output, there is some wobble you may allow at the input. Reliability everywhere is reliability at each setting, one at a time.

Continuity at $x$: the tolerance $V$ around $f(x)$ is demanded first, and the room $U$ around $x$ is found afterwards. The map has to send the whole of $U$ inside $V$. Order matters — a $U$ that works for a wide $V$ usually fails for a narrow one.

Let $f$ take $X$ to $Y$ and let $x\in X$. The map is continuous at $x$ when for every neighbourhood $$ of there is a neighbourhood of with . Since every neighbourhood contains an open one, it is enough to test open and produce open .

• A map is continuous exactly when it is continuous at every point, which the theorem below proves. So the two definitions never disagree, and each is used where it is convenient.
• In metric spaces the neighbourhoods can be taken to be balls, and the condition becomes the familiar one: for every $\varepsilon>0$ there is $\delta>0$ with $d(x,y)<\delta\implies d(f(x),f(y))<\varepsilon$.

For one direction take a point and a neighbourhood of its value: the preimage of an open set inside that neighbourhood is open and contains the point, so it is the room required. For the other, take an open set of the target and any point of its preimage: continuity at that point gives room around it that lands inside the open set, so the room lies inside the preimage. Every point of the preimage has an open set around it inside, which makes the preimage open.

If the map is continuous, the preimage of a neighbourhood of the value is open and holds the point.

That preimage is the room the local condition asks for.

For the converse take an open set of the target and a point of its preimage.

Continuity at that point gives room landing inside V, and such room lies inside the preimage.

Every point of the preimage has an open set around it inside, so the preimage is open and the map is continuous.

Continuity at $x$ demands: whatever neighbourhood $V$ of $f(x)$ is named, some neighbourhood $U$ of $x$ is sent inside it.

In metric spaces the neighbourhoods may be taken to be balls, and the local condition turns into the definition from the real line.

• Continuity at $x$: for every neighbourhood $V$ of $f(x)$ there is a neighbourhood $U$ of $x$ with $f(U$.

Great work! You now know how to:

• state continuity at a point;
• recover the $\varepsilon$–$\delta$ condition from it;
• prove that continuity everywhere is continuity at each point;
• give a map continuous at one point only.