# the closed interval [0,1] ( with the usual topology) [duplicate]

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, amWhyOct 17 '13 at 12:54

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.
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