Constructing a Piecewise Continuous Function at a Single Point

[gaborfk]

Homework Statement

For each $$a\in\mathbb{R}$$, find a function $$f$$ that is continuous at $$x=a$$ but discontinuous at all other points.

The Attempt at a Solution

I guess I am not getting the question. I need to come up with a function, I was thinking of a piecewise defined one, half rational half irrational, which is continuous on one but not the other? Is this possible?

Thank you in advance

 

[sutupidmath]

what about f(x)={0, x-rational, x, where x irrational.
take a sequence {a} that converges to 0, from this sequence let's take two subsequences {b} of rationals, and {c} of irrationals, since {a} converges to 0 also {b} and {c} should converge to zero. now let's take the corresponding sequence of the function

f({a})-->0

f({b})-->x-->0

So this function i guess is continuous at x=0, since also f(0)=0, but it is discontinuous everywhere else.

Let's see what other guys have to say on this, cuz, i am not 100% sure that what i did actually works.

 

[gaborfk]

Thank you!

That sound great.

 

[sutupidmath]

Thank you!

That sound great.

Can you show why the function that i took as an example, from the top of my head, is everywhere else discontinous, because i left this part for you to show.?

 

[gaborfk]

Because there are infinitely many irrational numbers which would make the graph continuous on the irrationals, but on an interval there would be rationals mixed in between the irrationals?

 

[sutupidmath]

Because there are infinitely many irrational numbers which would make the graph continuous on the irrationals, but on an interval there would be rationals mixed in between the irrationals?

Well, try to use the same logic i used to show that it is continuous at 0. In other words try to use sequences and see if you can come up with sth. It is quite trivial frome here, i guess.