Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Yesterday, user t.b. linked me a passage in Counterexamples in Topology.

In example 75, linked above, I have a questions on properties 2 and 5.

The closure of each basis neighborhood $N_\epsilon((x,y))$ contains the union of the four strips of slope $\pm\theta$ emanating from $B_\epsilon(x+y/\theta)$ and $B_\epsilon(x-y/\theta)$.

For property 2, why is the closure of a basis neighborhood the union of those 4 strips? After that I understand why any two basis neighborhoods must intersect.

Every real-valued continuous function $f$ on $(X,\tau)$ is constant, for if $f$ were not constant, $f(X)$ would contain two disjoint open sets with disjoint closures. The inverse images would then be disjoint open sets with disjoint closures, which is impossible.

For property 5, why does the image of a nonconstant continuous function contain two disjoint open sets with disjoint closures? Since the function is continuous, I understand the inverse image of the disjoint open sets will be disjoint, but why are the closures of the inverse images also disjoint? (Essentially, is there a more explicit reason than the one Steen and Seebach give?)

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

(2) Take a point $(x_0,y_0)$ in the union of the four strips, and let $N=N_\delta((x_0,y_0))$ be any basic nbhd of it. If it’s in one of the strips with negative slope, the ‘right foot’ of $N$ must intersect one of those intervals on the $x$-axis, and if it’s in one of the strips with positive slope, the ‘left foot’ of $N$ must do so. Thus, every nbhd of $(x_0,y_0)$ must intersect at least one of those two intervals, and $(x_0,y_0))$ must be in the closure of $N_\epsilon((x,y))$.

(5) Suppose that $f:X\to\mathbb{R}$ is not constant, and let $a$ and $b$ be distinct points in $f[X]$. Without loss of generality assume that $a<b$. Let $r=(b-a)/3$, let $U_a = (a-r,a+r)\cap f[X]$, and let $U_b=$ $(b-r,b+r)\cap f[X]$; then $U_a$ and $U_b$ are non-empty disjoint open subsets of $f[X]$, and their closures in $f[X]$ are $[a-r,a+r]\cap f[X]$ and $[b-r,b+r]\cap f[X]$, which are certainly disjoint. The inverse images under $f$ of disjoint sets are disjoint, so $f^{-1}[U_a]$ and $f^{-1}[U_b]$ are disjoint non-empty open sets in $X$ with disjoint closures.

share|improve this answer
    
Thanks Brian M. Scott. –  Danielle Intal Oct 22 '11 at 10:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.