Is the Inverse of a Continuous Function Always Continuous?

[seed21]

Homework Statement

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].<br /> Then, f[/itex] is continuos, if given (x_n) that converges at x \in X, then f((x_n))[/itex] converges at f(x)[/itex].&amp;amp;lt;br /&amp;amp;gt; I need a counter example, to prove that the reciprocal is not true.&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; All I know is that X should not be first countable.&amp;amp;lt;br /&amp;amp;gt; Please, help me.&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; Thanks in advance.

 

[HallsofIvy]

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 continuous at some number x in the real line. Giving different formulas to rational and irrational x might be useful.

 

[seed21]

That`s true.

Thanks for help.