KC implies US

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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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Any KC-space is a US-space.

Definitions used


Further information: KC-space

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


Further information: US-space

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


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_ns that are equal to y. This new sequence of x_ns converges to both x and y. Let A be the union of \{ x \} and the x_ns in this new sequence.

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

The last step contradicts our assumption that the x_ns converge to y.