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Let $X,Y$ be topological spaces and $f: X \to Y$. I know that if $X,Y$ are not necessarily first countable (=countable nbhood base) then

''For all sequences $x_n\to x$ in $X$ it's true that $f(x_n) \to f(x)$ in $Y$''

does not imply that $f$ is continuous.

I am trying to find an example of $f,X,Y$ such that for all sequences $x_n\to x$ in $X$ it's true that $f(x_n) \to f(x)$ in $Y$ but $f$ is discontinuous.

My first idea is $X=\mathbb R$ with the usual topology and $Y=\mathbb R$ with the discrete topology and then $f(x) = x$. Then $f$ is not continuous but also $f(x_n) \not\to f(x)$ even if $x_n \to x$. I am trying to think of spaces that do not have countable nhood bases but all I can think of is the discrete topology in an uncountable set.

Could anybody help me with an example?

Martin Sleziak
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Frosty
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  • IIRC, the usual example uses the first uncountable ordinal with the order topology. – egreg Apr 08 '14 at 12:07
  • Related question: http://math.stackexchange.com/questions/53236/continuity-and-image-of-convergent-sequences – Martin Sleziak Apr 08 '14 at 13:35
  • If you want to find such an example, $X$ needs to be a space which is not sequential. Good introduction into sequential spaces is at Dan Ma's topology blog. If you browse through the posts on this topic, you will find also some examples of spaces that are not sequential. – Martin Sleziak Apr 08 '14 at 13:42

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One of the simplest examples is using $X = \mathbb{R}$ in the co-countable topology (i.e. the closed sets are $X$ and the countable (including finite and empty) sets), and $Y$ as $\mathbb{R}$ in the usual topology, and as a map $f(x) = x$, the identity.

This function is not continuous (an open interval in the usual topology is not open in co-countable topology, as its complement is not countable), but $X$ has the property that any sequence $(x_n)$ in it that converges to some $x$ is eventually constant, i.e. for some $N$, all $x_n$ with $n > n$ are equal to $x$.

This follows from considering $A = \{x_n: x_n \neq x\}$ and noting that $O = X\setminus A$ is an open neighbourhood of $x$ and so all terms must be eventually in $O$....

So for any convergent sequence $(x_n)$, the image sequence under $f$, i.e. the same sequence but considered in $Y$, is also convergent (as eventually constant sequences are convergent to that constant in any topology, by definition). So $f$ is (trivially) sequentially continuous but as said, not continuous.

Henno Brandsma
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