Open/compact sets in metric $d_1$ and $d_2$ In a set $X$ we consider two metrics $d_1$ and $d_2$.
We consider that the identity map $f:(X,d_1)\rightarrow (X,d_2)$ with $f(x)=x$ is continuous.
Which of the following statements are correct?
(a) If a set is open as for $d_2$ then it is also open as for $d_1$.
(b) If a set is compact as for $d_2$ then it is also compact as for $d_1$.
(c) A set is open as for $d_1$ iff it is open as for $d_2$.
(d) If a set is open as for $d_1$ then it is open as for $d_2$.
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I think (d) is correct because for the other statements we would need also that $f$ is surjective, or not?
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The definition of continuity is:
A function $f:(X,d)\rightarrow (Y,\sigma )$ between two metric spaces is continuous in a point $x_0\in X$ if for each $\epsilon>0$ there is a $\delta>0$ such that $d(x,x_0)<\delta \Rightarrow \sigma (f(x),f(x_0))<\epsilon$. A function is continuous in each element iff the inverse image of an open (closed) set is open (closed).
 A: The first statement $(a)$ is true. To see this, let $U$ be a set open for $d_2$. Then, consider the the preimage $f^{-1}(U)$.
Since $f$ is continuous, we must conclude that because $U$ is open, $f^{-1}(U)$ is open in $d_1$. But, because $f$ is the identity map, we conclude that $f^{-1}(U) = U$.
The other statements are incorrect. To see this, consider the following counter-example:
Let $X = \mathbb{R}$, and suppose that $d_2$ is the standard metric (i.e. $d_2(x,y) = |x-y|$). Let $d_1(x,y) = \begin{cases} 0 \ \text{if } x = y\\ 1 \ \text{if } x \neq y\end{cases}
$
It isn't too difficult to show that $d_1$ is a metric on $\mathbb{R}$ AND that with these metrics, the identity map $f$ as you prescribed is continuous.
Now, pick a point $x \in \mathbb{R}$, and consider the ball
$B = \{y \in \mathbb{R}: d_1(x,y) < 1\}$. It isn't too hard to see that $B$ is precisely the singleton $\{x\}$ which is open in the topology induced by the metric $d_1$. However, singletons are NOT open in the standard euclidean topology. That is, $\{x\}$ is not open in $d_2$. This immediately rules out statements $(c)$ and $(d)$.
As far as why $(b)$ isn't true, we note that the closed interval $I = [0,1]$ is compact in $d_2$. However, it cannot be compact in $d_1$. Why not? Well, $A = \{ \{x\}\}_{x \in I}$ is an open cover of $I$ (in the metric topology induced by $d_1$) which admits no finite subcover.
Edit: To answer some of the questions in the comments, I've included this edit. Let's start with why singletons $\{x\}$ are open in the topology induced by $d_1$. Remember that when we have a metric $d$ on a space $X$, we define the metric topology on $X$ induced by $d$ as the topology generated by balls. That is, any set taking the form $\{y \in X: d(x,y) < r\}$ for some $x \in X$ and some $r > 0$ is automatically open, by definition. So, looking back at the set $B$ which I defined in my original answer, we see that the set $B = \{y \in \mathbb{R}: d_1(x,y)<1\}$ takes precisely this form, so it is open, by definition of the metric topology induced by $d_1$. However, we claim that $B = \{x\}$. Why? Well, clearly $d_1(x,x) = 0$ (recall the axioms for a metric), so $x \in B$ by definition of $B$. However, if $y \neq x$, then our definition of $d_1$ tells us that $d_1(x,y) = 1$. Since $d_1(x,y)$ is NOT strictly less than $1$, we conclude that $y$ cannot be in $B$, so it follows that the only element of $B$ is precisely $x$ itself. Hence we have showed that $B$ is open (trivially) and that $B = \{x\}$.
To answer your next question: Why are singletons not open in the standard Euclidean topology? In general, we have to think back to the definition of an open set in a metric topology. We can say the following: Let $(X,d)$ be a metric space. A subset $U \subset X$ is open (with respect to the metric topology induced by $d$) if for each $x \in U$ there is a ball $V$ containing $x$ satisfying $V \subset U$. With this definition, it isn't too hard to see why singletons can't be open in the standard Euclidean topology on $\mathbb{R}$: Any ball $V$ containing the point $x \in \mathbb{R}$ cannot be a subset of the singleton $\{x\}$ (balls in the standard Euclidean topology are precisely open intervals on the real line, but no open interval is a subset of a singleton).
Next question: Why is $I = [0,1]$ compact in $d_2$, but not $d_1$? A simple way to see that $I$ is compact (in $d_2$) is by invoking the Heine-Borel Theorem:

A subset of $\mathbb{R}$ is compact (in the standard topology) iff it is closed and bounded.

Why is $I$ bounded? Well, $d_2(0,y) < 2$ for all $y \in I$ (i.e. there is a ball, centered at $0$ in our case, which contains $I$). I leave it as an exercise for you to show that $I$ is closed. Depending on which definition of closed you are working with, you may either have to show that the complement of $I$ (that is $\mathbb{R} \setminus I = (-\infty, 0) \cup (1, \infty))$ is open using the definition I gave above, or you may have to show that $I$ contains its limit points (being closed has many equivalent definitions). Ultimately, it ends up being a standard fact that all closed intervals in $\mathbb{R}$ are compact.
While the Heine-Borel theorem gives us a nice way to view compact subsets of Euclidean spaces, compactness can be defined more generally for arbitrary metric spaces $(X,d)$ (I limit myself to metric spaces since you have this question posted under the analysis tag) in the following way:
Let $(X,d)$ be a metric space. An open cover of $X$ is a family $A = \{A_j\}_{j \in J}$ of open subsets of $X$ (i.e. each $A_j \subset X$ is open) satisfying $$X \subset \bigcup_{j \in J} A_j$$
Here, $J$ is just an arbitrary indexing set. Then, $X$ is compact if every open cover $A$ of $X$ has a finite subcover (i.e. if $A$ is an open cover of $X$, then there exists a finite subset $B \subset A$ such that $B$ is an open cover of $X$).
With these definitions, it is easy to see that the set $A = \{\{x\}\}_{x \in I}$ which I defined in my original answer is an open cover of $I$ (in $d_1)$. In $d_1$, we proved that singletons are open, so $A$ consists of open subsets of $I$. Also, $$I = \bigcup_{x \in I} \{x\}$$
Hence, $A$ satisfies the definition of an open cover of $I$. However, no finite subcover exists. Why? For a contradiction, suppose a finite subcover $B = \{\{x_i\}\}_{i=1}^n \subset A$ exists. Then, by definition of an open cover, $$I \subset \bigcup_{i=1}^n \{x_i\}$$
But this inclusion is impossible since $I$ contains infinitely many elements while the set on the right hand side of the inclusion is finite. Thus, no finite subcover of $A$ exists, hence $I$ cannot be compact in $d_1$.
