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**Definition**

If *A* is a subset of a metric space *X* then *x* is a **limit point of A** if it is the limit of an eventually non-constant sequence (

**Remarks**

- This is the most common version of the definition -- though there are others.

- Limit points are also called
*accumulation points*.

**R**with the usual metric

Sets sometimes contain their limit points and sometimes do not.

- The points 0 and 1 are both limit points of the interval (0, 1).

- The set
**Z****R**has*no*limit points.

For example, any sequence in**Z**converging to 0 is eventually constant.

- The points 0 and 1 are both limit points of the interval (0, 1).
**R**^{2}with the usual metric

The limit points of the*open*disc {(*x*,*y*)**R**^{2}|*x*^{2}+*y*^{2}< 1} form the*closed*disc {(*x*,*y*)**R**^{2}|*x*^{2}+*y*^{2}1}.

Any point on the boundary of the circle is a limit of a sequence of points inside the circle.- In
**R***every*real number is a limit point of the subset**Q**of rationals.

**Proof**

Every real number can be approximated arbitrarily closely by a sequence of rationals (by truncating the decimal expansion, say).

Alternative definitions of a limit point of a subset

- A point
*x*of*X*is a limit point of*A*if every -nd of*x*in*X*meets*A*in a point*x*.

- A point
*x*of*X*is a limit point of*A*if every open set of*X*containing*x*meets*A*in a point*x*.

We can connect this idea with the last section by:

A subset

**Examples**

- The subset
*X*is a closed subset of itself. The empty set is closed. Any*finite*set is closed.

- The closed interval [0, 1] is closed subset of
**R**with its usual metric. There are, however, lots of closed subsets of**R**which are not closed intervals.

- The closed disc, closed square, etc. are closed subsets of
**R**^{2}.

Then the connection with the last section is:

**Theorem**

*A subset A of a metric space X is closed if and only if its complement X - A is open.*

**Proof**

() Suppose A is closed. We need to show that *X* - *A* is open. So suppose that *x* *X* - *A*. Then some -neighbourhood of *x* does not meet *A* (otherwise *x* would be a limit point of *A* and hence in *A*). Thus this -neighbourhood of *x* lies completely in *X* - *A* which is what we needed to prove.

() Conversely, suppose that *X* - *A* is open. We need to show that *A* contains all its limit points. So suppose *x* is a limit point of *A* and that *x* *A*. Then *x* *X* - *A* and hence has an -neighbourhood *X* - *A*. But this is an -neighbourhood that does not meet *A* and we have a contradiction.

**Recap**

The last two sections have shown how we can phrase the ideas of continuity and convergence purely in terms of open sets.

- A map
*f*:*X**Y*is continuous*B*open in*Y**f*^{-1}(*B*) is open in*X*.

- A point
*x*is a limit point of a subset*A*of*X*every open set containing*x*meets*A*(in a point*x*).

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