Is the following set open/closed/compact in the metric space?

So I just had a metric space examination today and I want to see if I answered this question correctly.If someone could point out if I went wrong and maybe tell me how many marks you think I might have gotten out of the marks the questions are worth ?

consider the metric space $(X,d)$ where $X=[0,1]$ is the set of functions $f:[0,1] \rightarrow [0,1]$, and the distance $d$ is given by

$d(f,g)=||f-g||_\infty=sup\{|f(t)-g(t)| \Bbb|t \in[0,1] \}$

let

$J= \{f \in X | \forall x,y \in [0,1], x \le y :f(x) \le f(y) \}$

be the set of non-decreasing functions in $X$

(a) is $J$ closed in $(X,d)$ (12 marks)

(b)is $J$ open in $(X,d)$ (12 marks)

(c) is J compact (12 marks)

(a) In order to decide if $J$ is closed , we see if $J^C$ is open or not.

$J^C$ is the set of all strictly decreasing functions s.t. if $x<y \Rightarrow f(x)<f(y)$ also note that $f(x)$ is bounded below by $0$ and above by $1$

in order for $J^C$ to be open there must exist some $B_\varepsilon(f(x)) \subset J^C \Rightarrow d(f(x),g(y))< \varepsilon$

$d(f(x),g(x)=sup\{|f(t)-g(t)| \}$ so we seek to find an $\varepsilon$ s.t. $|f(t)-g(t)|< \varepsilon \forall t \in [0,1]$

clearly any $\varepsilon$ greater than one satisfies this property so $J^C$ is open $\Rightarrow J$ closed.

(b)$J$ is the set of non-decreasing functions If $J$ is to be open $\Rightarrow \exists B_\varepsilon(f(x)) \subset J \Rightarrow sup\{|f(t)-g(t)| \} < \varepsilon$ i.e. $|f(t)-g(t)| < \varepsilon \forall t \in [0,1]$

as they are non decreasing and bounded $\exists M_1, M_2 \in \Bbb R$ s.t. $f(t)\le M_1$ and $g(t) \le M_2 \forall t \in [0,1] \$

$|f(t)-g(t)| \le |f(t)| +|g(t)| \le M_1 + M_2$ now if we let $M_1+M_2= \varepsilon \Rightarrow |f(t)-g(t)| \le \varepsilon$ and so is not open as this is a closed ball.

(c) by the Heine Borrel theorem as J is closed and bounded it is also compact .

• $J^c$ is not the set of all strictly decreasing functions. It consists of an the functions that are not non decreasing over the entire interval $[0,1]$. For example it contains functions that are non decreasing on $[0,1/2)$ and non increasing on $(1/2,1]$ – Abishanka Saha Jan 13 '18 at 16:36

(a) No, $J^c$ is not the set of strictly decreasing functions! Most functions are neither increasing nor decreasing, but every function must be in $J$ or $J^c$.

In fact $J^c$ is the set of functions $f$ such that there exist $x,y$ with $x<y$ and $f(x)>f(y)$.

In fact $J^c$ is open, but the set of strictly decreasing functions is not. So your proof that it's open must be wrong. In fact that proof makes no sense at all, because you don't ever mention "strictly decreasing" - if that proof were right it would be a proof that every set is open.

In fact to show $J^c$ is open you need to show this: If $f\in J^c$ then there exists $\epsilon>0$ so that if $d(f,g)<\epsilon$ then $g\in J^c$. Showing that is not at all the same as showing that $d(f,g)$ is less than $\epsilon$.

(b) Similar confusion. Saying $B_\epsilon(f)\subset J$ does not imply that $d(f,g)<\epsilon$, as you say. It says that if $d(f,g)<\epsilon$ then $g\in J$.

(c) The Heine-Borel theorem applies to subsets of $\Bbb R^n$. Since $J$ is not a subset of $\Bbb R^n$ you can't use it here.

Be prepared for bad news: If I were grading this you'd get a 0 on the whole thing. Probably I'd add a note that it looks like you're really not ready for this class - first you should take a more elementary "proof class", to get straight how you verify if-then statements, that sort of thing.