[passer-by]

6. By definition of f , then we have

$ \mid f(x)-g_{\frac{\epsilon}{3}}(x) \mid < \frac{\epsilon}{3} $ ...(1)
And $ \mid f(y)-g_{\frac{\epsilon}{3}}(y) \mid < \frac{\epsilon}{3} $...(2)

Since $g$ is uniformly continuous on A ,then there exists $ \delta >0 $ such that

$ \mid g_{\frac{\epsilon}{3}}(x)-g_{\frac{\epsilon}{3}}(y) \mid < \frac{\epsilon}{3} $ (if $ \mid x-y \mid < \delta $) ...(3)

So if we combine (1) to (3) then we have

$\mid f(x)-f(y) \mid \leq \,\mid f(x)-g_{\frac{\epsilon}{3}}(x) \mid + \mid g_{\frac{\epsilon}{3}}(x)-g_{\frac{\epsilon}{3}}(y) \mid + \mid g_{\frac{\epsilon}{3}}(y)-f(y) \mid < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} =\epsilon $(if $ \mid x-y \mid < \delta $ and all x,y in A)

Hence, f is uniformly continuous on A.

7. Neither open nor closed set.

Let $ x \in Q$

Since every neighborhood of x contains irrational numbers ,then Q can't be open set.

Moreover, we can construct sequence of points in Q converging to irrational number,say ,y , so y is a limit point of Q, but not in Q. It follows that Q isn't closed set.

NOTE: In mathematics, set seems different from door. Sometimes it's both closed and open simultaneously and sometimes it's neither closed nor open simultaneously.

8. No.

For example, $ K_n= [n,n+1] $ which is compact in R. So $$ \bigcup_{n=1}^{\infty}=[1,\infty)$$ Such countable union isn't compact (since it's unbounded in R).


9. (I 'll do this problem based on compact in R only)

Since compact implies bounded ,then inf K and sup K exist.

The remaining is to prove that both are in K. We'll prove for sup K only and inf K is done in similar way.

Using characterization of supremum, for all $\epsilon >0$ there exists $x \in K $ such that
$\text{sup K} - \epsilon < x $...(1)

Since $ x \leq \, \text{sup K} < \text{sup K} +\epsilon $...(2)

Combine (1),(2) we obtain$ \mid \text{sup K} - x \mid < \epsilon $

Equivalently, all neighborhoods of sup K contains element in K and thus sup K is a limit point of K.

But K is compact , so it's closed and hence K contains all limit points,including sup K.

10. Let $ x,y \in I $ and suppose that x<y

By Mean Value Theorem of f on (x,y),

so there exists $ c \in (x,y) $ such that $\frac{f(y)-f(x)}{y-x} = f'(c) < 0 $

Thus, f(y)-f(x) < 0 ,that is , f is strictly decreasing.