KC implies US

This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., KC-space) must also satisfy the second topological space property (i.e., US-space)
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Statement

Any KC-space is a US-space.

Definitions used

KC-space

Further information: KC-space

A topological space is termed a KC-space if every compact subset of it is closed.

US-space

Further information: US-space

A topological space is termed a US-space if every convergent sequence has a unique limit.

Proof

Given: A KC-space $X$ .

To prove: If $\{x_{n}\}_{n\in \mathbb {N} }$ is a sequence with limit $x$ and limit $y$ , then $x=y$ .

Proof: Clearly, $x_{n}$ cannot be eventually constant at both $x$ and $y$ . Let us say that it is not eventually constant at $y$ . Throw out all the $x_{n}$ s that are equal to $y$ . This new sequence of $x_{n}$ s converges to both $x$ and $y$ . Let $A$ be the union of $\{x\}$ and the $x_{n}$ s in this new sequence.

$A$ is compact: Any open cover of $A$ contains one open set containing $A$ , so it contains all but finitely many of the $x_{n}$ s. Thus, that along with finitely many other open subsets covers $A$ . Thus, every open cover has a finite subcover.
$A$ is closed: This follows from the previous step, and the assumption that $X$ is a KC-space.
There is an open subset containing $y$ that does not contain any of the $x_{n}$ s: This follows from the fact that $A$ is closed, and $y\notin A$ .

The last step contradicts our assumption that the $x_{n}$ s converge to $y$ .