A sequence converges to a point when its terms end up inside every ball around that point, however small the ball is asked to be. Only the distance is used, so the definition reads the same in the plane, in a space of functions and in a set with the discrete metric — and the answers differ wildly.

A plane approaching an airport is eventually inside every circle you care to draw around it: within ten miles, then within one, then within a hundred metres. The circles are yours to choose; the aircraft has to satisfy all of them.

The terms of $x_n=1/n$ against their index, with a band of half-width $\varepsilon=0.22$ about the limit $0$. From the fifth term on, every term is inside the band; that is what the $N$ marks. Shrink the band and the mark moves right, but a mark can always be found — which is exactly what convergence asks.

Let $(X,d)$ be a metric space and $x_1,x_2,x_3,\dots$ a sequence of its points. The sequence converges to when the distances can be forced below any tolerance by going far enough along the sequence. We write , and call the limit.

• The tolerance comes first and the index second: $N$ is allowed to depend on $\varepsilon$, and it always does. A smaller $\varepsilon$ buys a larger $N$.
• Every term from $N$ on must be inside the ball $B(x,\varepsilon)$. So convergence says: each ball about contains all but finitely many terms of the sequence.

Suppose the sequence had two different limits. Their distance apart is a positive number, so half of it is a tolerance the definition must meet for both of them at once. Beyond some index every term is within that half of the first limit and within that half of the second, and then the triangle inequality makes the distance between the two limits smaller than itself. That is impossible, so the two limits were never different.

Assume two different limits and take half their distance apart as the tolerance; it is positive because the points differ.

The definition applied to the first limit gives an index beyond which the terms are that close to x.

The same definition applied to the second limit gives another index.

Take any n beyond both indices and go from x to y through that term.

But twice the tolerance is the distance itself, so the distance is smaller than itself. The assumption that the limits differ has to go.

Convergence asks for one thing for every positive $\varepsilon$: an index beyond which the terms stay within $\varepsilon$ of the limit.

Take $\varepsilon=\tfrac12$. The only points within $\tfrac12$ of are itself, so the terms beyond must all equal .

• $x_n\to x$ means: for every $\varepsilon>0$ there is an $N$ with $$ for all .

Great work! You now know how to:

• state convergence in a metric space and read its quantifiers in order;
• find an $N$ for a given $\varepsilon$;
• use that a limit is unique and depends only on the tail;
• say what convergence means under the discrete metric.