5

This question is a subproblem of the A map is continuous if and only if for every set, the image of closure is contained in the closure of image.

I am not able to come up with any example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$. Can anyone give such an example?

José Carlos Santos
  • 427,504
MUH
  • 1,377
  • Sorry, I found two instances of this same question in the "Related" section. I am just referring to them. (1) and (2) – MUH Mar 11 '18 at 12:27

4 Answers4

3

Hint : consider $f = \arctan$ and $A = \Bbb R$.

krirkrirk
  • 2,027
  • Ah, I was trying to think in the same line but couldn't come up with the examples that $\overline{A}$ should include such values that doesn't add any new points in $y$ values of the function but the closure of $y$ values does add. – MUH Mar 11 '18 at 12:16
  • Basically, any function $f$ that is continuous on $\Bbb R$, but such that $f(\Bbb R) \neq \Bbb R$ and is open, will do the trick. – krirkrirk Mar 11 '18 at 12:24
3

Take $f\colon\mathbb{R}\longrightarrow\mathbb R$ define by $f(x)=\frac1{1+x^2}$ and $A=\mathbb R$. Then $\overline A=\mathbb R$, $f(A)=(0,1]$, and $\overline{f(A)}=[0,1]$.

Another example is the identity function $f$ from $\mathbb R$ endowed with the discrete metric into $\mathbb R$ endowed with the usual metric. Take $A=\mathbb Q$. Then $\overline A=\mathbb Q$ (with respect to the discrete metric). But $\overline{f(\mathbb{Q})}=\overline{\mathbb{Q}}=\mathbb R$ (with respect to the usual metric).
José Carlos Santos
  • 427,504
  • This example is also on the same line as above. Thanks. – MUH Mar 11 '18 at 12:18
  • 1
    @MUH I've added another example along a different line. – José Carlos Santos Mar 11 '18 at 12:23
  • The latter example is a nice one. I just wanted to check that the function is continuous. Let $V \in \mathbb{R}$ endowed with usual metric be any open set in topological space corresponding to Euclidean metric, then $f^{-1}(V) = { q \in \mathbb{Q} | q \in V }$. Now this set is also open in $\mathbb{R}$ endowed with discrete metric. – MUH Mar 11 '18 at 12:48
  • 1
    @MUH If a set $S$ is endowed with the discrete metric, then every function $f$ from $S$ into any metric space $M$ is continuous. This is so because, if $A$ is an open subset of $M$, then $f^{-1}(A)$ is open (because every subset of $S$ is open). – José Carlos Santos Mar 11 '18 at 12:51
  • One off-topic question. Do you have these results in mind just because of practice or does it come naturally? I mean I can derive these results (it is quite obvious here) but since I am studying some other maths also, I end up doing the same thing again and again but still don't remember the results (not that it is important but still) – MUH Mar 11 '18 at 12:59
  • @MUH In the case of both of my answers to your question, your first suggestion is the right one: I have these results in mind just because of practice. – José Carlos Santos Mar 11 '18 at 13:04
1

Let $f:X \to Y$ be any continuous, non-surjective function with dense range in $Y$. Then $$f(\overline{X}) = f(X) \neq \overline{f(X)} = Y$$

An explicit example is given by $f: \ell^2 \to \ell^2$ defined by $f(x) = (2^{-n} x_n)_{n \geq 1}$ since the space $c_{00}$ is contained in the range of $f$ and hence the range of $f$ is dense but e.g. the sequence $(2^{-n})_{n \geq 1}$ is easily seen to not be in the range of $f$.

Rhys Steele
  • 19,671
  • 1
  • 19
  • 50
0

Let $f:X\to Y$ where $Y$ is equipped with indiscrete topology.

Then automatically $f$ is continuous and $\overline{f(A)}=Y$ for every non-empty $A\subseteq X$.

For suitable $X$ plenty examples can be found with $\varnothing\neq A\subseteq X$ and$f(\overline{A})\neq Y$.

drhab
  • 151,093