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A topological space is called a US-space provided that each convergent sequence has a unique limit.

A topological space is called $T_B$ if each compact subset is closed.

The bellow example show that $US$ space is not $T_B$ space.

Suppose X with the closed interval [0,1] ( with the usual topology) and attach a new point $z$ whose neighborhoods are open dense subsets of [0,1].[0,1] is a compact non closed subspace of X and thus X is not $T_B$ space. However no sequence in [0,1] converges to $z$ and in particular sequences in X have unique limits.

Why is it right "no sequence in [0,1] converges to $z$ and in particular sequences in X have unique limits"?

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marked as duplicate by Cameron Buie, Davide Giraudo, Lord_Farin, azimut, amWhy Oct 17 '13 at 12:54

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1 Answer 1

Recall the definition of convergence:

A sequence $\{x_n\}\in X$ converges to $x \in X$ iff for any open neighbourhood $U \ni x$ there exist an $N$ such that $x_i \in U$ for all $i > N$.

If a sequence entirely contained in $[0, 1]$ is to converge toward $z$, then for any neighbourhood of $z$, we must have that eventually the sequence is contained in that neighbourhood. But this isn't the case. I can construct a neighbourhood for which this fails. If the sequence is contained in $[0, 1]$, then it has an infinite subsequence whose complement $U \subset [0, 1]$ is dense and open, and thus $\{z\}\cup U$ is an open neighbourhood of $z$, and the sequence will not eventually be contained there. Therefore the sequence does not converge toward $z$.

As for why sequences in $X$ have unique limits, we can say, with a very similar argument to the above, that if a sequence is to converge to $z$, it must eventually be constant and equal to $z$ (otherwise you can find a neighbourhood around $z$ which the sequence will leave at arbitrarily high indexes). If it does not converge to $z$, but some other number $x \in [0, 1]$, then there is a neighbourhood around $X$ that does not contain $z$. Once the sequence is contained in that neighbourhood (which will happen eventually, by the definition of convergence), then we can pretend $z$ doesn't even exist, and the result follows from the fact that limits are unique on the number line.

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The complement of the entire sequence won't necessarily work, as was David's point. Rather, we use the fact that $[0,1]$ is a sequentially compact subspace to get a subsequence that converges to a point $x$ of $[0,1].$ Then the union of the points of that subsequence and the singleton $\{x\}$ is closed and nowhere dense in $[0,1],$ and its complement is the desired neighborhood of $z$. –  Cameron Buie Oct 17 '13 at 12:25
    
@CameronBuie Apparently I hadn't removed all hints of the flaw from my answer as I fixed it. I believe that it is done now. –  Arthur Oct 17 '13 at 12:29

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