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Can somebody tell me why every sequence in X converges to every point of X if we consider the indiscrete topology $\tau=${$\emptyset,X$}?

Roos Jansen
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  • Do you know what it means for a sequence to converge in a general topological space? – user642796 Jun 25 '14 at 12:51
  • Yes, a sequence (xn) converges to x if for every neighborhood V of x there exists an integer k such that (xn) is in V for all n>=k. – Roos Jansen Jun 25 '14 at 12:52
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    So, here, there would be only one nhood of a given $x$... – David Mitra Jun 25 '14 at 12:54
  • I don't get it.. – Roos Jansen Jun 25 '14 at 12:58
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    Okay, so pick an arbitrary point $x$ of the indiscrete space $X$, and pick an arbitrary open neighbourhood $V$ of $x$. Now we want to find a $k$ such that $x_n \in V$ for all $n \geq k$. ... But wait a minute: $X$ is indiscrete, so what are the possible choices for $V$? – user642796 Jun 25 '14 at 13:01

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The definition of convergence in topology spaces is: Let X be a topology space and a sequence given by $(x_1,x_2,x_3,...) \in X^{\mathbb N}$. A point $x \in X$ is called limit point of $(x_n)_{\mathbb N}$ if for every open set U $\in X$ which contains x holds: There exists an index $n_0 \in \mathbb N$ such that $x_n \in U$ $\forall n \geq n_0$.

So in our example we only have one open set which can contain a point $x \in X$, namely X. This is the only (not-empty) open set in our topology space, hence for every limit point in X we got the same open set. And this set contains all points of X. Thatswhy every sequence converges.

Marm
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