Continuous function h:R->R and K={x:h(x)=0} is closed

[felvt]

Given a function h: R --> R is continuous on R, let K={x: h(x)=0}. Show K is a closed set.

Not sure how to do this, but I think I am supposed to use the epsilon-delta definition of the continuous function and relate that to the definition of a closed set being a set that contains all of its limit points using the epsilon-nieghborhood definition of the limit point. Might be an easier way to go about this though.

 

[daon]

If K is empty then its true, right? Now if K is not empty, let L = R-K. If we show L is open, then K is closed.

If K = R or K=R-{q1, q2, ..., qn} for some real numbers q1, ..., then we're done (the empty set and R minus a finite point set are open). So assume L contains at least a countably infinite number of points.

Let p be any point in L. h(p) is not zero, right?

We are garenteed there is an open ball of radius |h(p)| around h(p) that contains no points on the x axis.

....i.e. if (t,0) is a point on the x axis, |h(p)-h(t)| = |h(p)| which means (t,0) is not in an open ball of radius |h(p)| around h(p).

Now, let q1 be the largest real number less than p such that h(q1)=0, and let q2 be the smallest real number greater than p such that h(q2)=0. These must exist by the continuity of h, since for all e>0 we can find an interval (p-d,p+d) such that if s is in (p-d,p+d) h(s) is closer than e to h(p)... i.e. for all s in (p-d,p+d), h(s) is not zero if we choose a small enough e.

the interval Q=(q1, q2) contains no elements of K and contains p.

Therefore, Q is contained entirely inside L, and is open. therefore p is contained inside an open set lying completely inside L.

Since the choice of p was arbitrary in L, L is open. Which means K is closed.

 

[PAULK]

I think you can do it directly:

Let x0 be any limit point of K. [To prove: x0 is in K]
Then there is a sequence x1, x2..., all points in K, that converges to x0.

Now h(xj) = 0 for all xj in K.

But h is continuous, so lim[xj -> x0] h(xj) = h(x0). But lim[xj -> x0] 0 = 0 = h(x0).

Therefore x0 is in K

 

[daon]

haha. I always had a knack for doing things in an awkward manner. Yours, Paul, is far more straight forward.