Im trying to solve a problem from chapter 8, Real Analysis, Carothers, 1ed, talking about compactness of metric space, : enter image description here

I've finished the first problem actually.

For the second problem. I've proved the range of $f$ is closed with claim as following:

$f$ is continuous and $[0,1]$ is compact => range of f is compact in $[0,1]\times[0,1]$.

Since $[0,1]\times[0,1]$ is compact,

then we get range of $f$ is closed.

And then I got some problems:

  1. The metric is unknown, which means that it is insufficient to say $[0,1]$ is compact and so does $[0,1]\times[0,1]$. How to solve it?

  2. How to do next? I cannot use the hint left below.

  • $\begingroup$ metric on $[0,1]\times [0,1]$ would be the one induced from usual metric on $\mathbb{R}\times \mathbb{R}$ $\endgroup$ – user87543 Nov 29 '13 at 4:48
  • 1
    $\begingroup$ How did you prove the first problem? $\endgroup$ – Don Larynx Nov 29 '13 at 4:48
  • 2
    $\begingroup$ @DonLarynx: suppose that f can be onto. Then [0,1] and [0,1]*[0,1] are homeomorphic. Then I pick up a point from [0,1]*[0,1] and without this point, it is still connected while [0,1], missing a point corresponding to the point in square, is disconnected. Contradiction!(homemorphism) $\endgroup$ – Bear and bunny Nov 29 '13 at 4:53

You don’t really need a metric on $[0,1]\times[0,1]$; it’s more convenient to use the product topology directly. It has a base consisting of all sets of the form $I\times J$, where $I$ and $J$ have one of the following forms:

  • $(a,b)$ for $0\le a<b\le 1$;
  • $[0,b)$ for $0<b\le 1$; or
  • $(a,1]$ for $0\le a<1$.

Now let $K$ be the range of $f$; you already know that $K$ is closed, since it’s a compact subset of the Hausdorff space $[0,1]\times[0,1]$. Thus, it is nowhere dense if and only if its interior is empty. Suppose, on the contrary, that it contains a non-empty open set. Then it contains a basic open set of the form $I\times J$ described above, and sinec it’s closed, it contains the closure of that basic open set; this is a rectangle $[a,b]\times[c,d]$ for some $a,b,c,d$ such that $0\le a<b\le 1$ and $0\le c<d\le 1$.

To use the hint, show that $R=[a,b]\times[c,d]$ must be the image of some subinterval $I$ of $[0,1]$. Further HINT: If not, there are $u,v,w\in[0,1]$ such that $u<v<w$, $f(u),f(w)\in R$, and $f(v)\notin R$; now consider the sets $$R\cap f\big[[0,v]\big]\qquad\text{and}\qquad R\cap f\big[[v,1]\big]\;.$$

Then let $g=f\upharpoonright I$, the restriction of $f$ to the subinterval $I$; $g$ is continuous and maps $I$ onto the rectangle $[a,b]\times[c,d]$, but this is impossible for the same reason that $f$ cannot map $[0,1]$ onto $[0,1]\times[0,1]$.

  • $\begingroup$ What is Hausdorff space(Im sorry that I haven't learned about Topology)? $\endgroup$ – Bear and bunny Nov 29 '13 at 5:30
  • $\begingroup$ @Frank: A space $X$ is Hausdorff if for each $x,y\in X$ with $x\ne y$ there are open sets $U$ and $V$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$. You don’t have to worry about it, because every metric space is Hausdorff: just let $r=\frac12 d(x,y)$, where $d$ is the metric, and set $U=B(x,r)$ and $V=B(y,r)$. $\endgroup$ – Brian M. Scott Nov 29 '13 at 5:33
  • $\begingroup$ Why does it contain a basic open set of the form I×J? I mean the open set can be any shape. $\endgroup$ – Bear and bunny Nov 29 '13 at 5:37
  • $\begingroup$ @Frank: Because every open set in the plane contains an open rectangle. Do you know what is meant by product topology? $\endgroup$ – Brian M. Scott Nov 29 '13 at 5:40
  • $\begingroup$ Actually, I donnot know about the product topology. Wikipedia claimed that a product space is the cartesian product of a family of topological spaces equipped with a natural topology. Problem is I donnot know much about topology but its characteristic of no metric equipped. $\endgroup$ – Bear and bunny Nov 29 '13 at 5:45

If the range of $f$ has nonempty interior, then it contains a nonempty closed disk $D$. Choose points $x_0,x_1,x_2\in[0,1]\cap f^{-1}(D), x_0\lt x_1\lt x_2$. Let $y_1=f(x_1)\in D$. Let $A=D\cap f([0,x_1]),B=D\cap f([x_1,1])$. Then $A,B$ are closed sets, $A\cup B=D$, $A\cap B=\{y_1\}$. The sets $A\setminus\{y_1\},B\setminus\{y_1\}$ are disjoint, nonempty (containing the points $f(x_0),f(x_2)$ respectively), and relatively closed in $D\setminus\{y_1\}$, contradicting the fact that $D\setminus\{y_1\}$ is connected.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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