Two simple questions in topology - help:

Question

I would be happy with some advice with these problems:

1. Let $X$ be a compact space and $Y$ be a hausdorff space. $f:X\to Y$ a continuous map. Assume $F\subset Y$ is closed and $f^-1(F)\subset U\subset X$ be an open set of $X$. Prove there exists an open set $F\subset V\subset Y$ such that $f^-1(V)\subset U.$

2. Let $A\subset \mathbb R^n$ be a closed connected set. Prove that the set of points $x\in \mathbb R^n$ that satisfy ${d(x,A)\leq\epsilon }$ is also connected.

I have a difficulty with the following:

1.There seems to be no clear way to construct this V. $f(U)$ needs not be open (nor have some nice interior). It seems a higher seperation axiom is required.

2. If the question were about path connectivity, it is very easy and clear. However in this case, I don't really have an idea about how to use metricity here, and what strategy should I take to try and prove connectivity.

Answer 1

For the first one, consider $$V = Y \setminus f(X \setminus U)$$

First, $V$ is open : indeed, $X \setminus U$ is closed in the compact $X$, so it's compact, so its image $f(X \setminus U)$ is also compact because $f$ is continuous, so it's closed, so its complement in $Y$ is open.

Secondly, $F \subset V$ : indeed, let's suppose that you have $y \in F \setminus V$, then $y \in F \cup f(X \setminus U)$, so because $f^{-1}(F) \subset U$, this implies that $y \in F \cup f(X \setminus f^{-1}(F))$, which is absurd.

Finally, $f^{-1}(V) \subset U$ : indeed, if $x \in f^{-1}(V)$, then $f(x) \in V$, so $f(x) \notin f(X \setminus U)$, so $x \notin X \setminus U$, so $x \in U$.

Answer 2

For the second one, let's note $B = \lbrace x \in \mathbb{R}^n, d(x,A)\leq \varepsilon \rbrace$.

For all $x \in B$, let $c_x$ be the image of a path included in $B$ joining $x$ to $A$. Of course, $c_x \cup A$ is the union of two connected subsets that intersect so it is connected. Then, notice that $$B = \bigcup_{x \in B} (c_x \cup A)$$

So $B$ is a union of connected subsets that intersect (all of them contain $A$), so $B$ is connected.