# Counterexample Topology

1. Dec 7, 2008

### seed21

1. The problem statement, all variables and given/known data

Let $$( X, \tau_x)$$ $$(Y, \tau_y)$$ topological spaces, $(x_n)$ an inheritance that converges at $$x \in X$$, and let $$f_*:X\rightarrow Y[/itex]. Then, [tex]f[/itex] is continuos, if given $(x_n)$ that converges at [tex]x \in X$$, then [tex]f((x_n))[/itex] converges at [tex]f(x)[/itex].
I need a counter example, to prove that the reciprocal is not true.

All I know is that X should not be first countable.

Last edited: Dec 7, 2008
2. Dec 8, 2008

### HallsofIvy

Staff Emeritus
I think there are some translation problems here. $(x_n)$ is a "sequence" not an "inheritance". And you want to show that the "converse" of that statement, not the "reciprocal", is false.

The converse of "If for any sequence $(x_n)$ that converges to x, $(f(x_n))$ converges to f(x) then f is continuous at x" is "if f is continuous at x, then for any sequence $(x_n)$ that converges x, $(f(x_n))$ converges to f(x)".

I wonder if you don't have the statement reversed. The converse, as stated, IS true and there is no counter example.

However, if the original statement were "if f is continuous at x and $(x_n)$ is a sequence that converges to x, then $(f(x_n))$ converges to f(x)", its converse, "if $(x_n)$ is a sequence converging to x and $(f(x_n))$ converges to f(x), then f is continuous at x" is false. It might happen that there exist such a sequence (but other sequences,$(a_n)$ converging to x for which $(f(a_n))$ does NOT converge to f(x)) but f(x) is not continuous at x.

To look for a counter example, an obvious thing to do is to look at functions that are NOT continous at some number x in the real line. Giving different formulas to rational and irrational x might be useful.

3. Dec 8, 2008

### seed21

That`s true.

Thanks for help.