Linked Questions

7
votes
5answers
5k views

$f\colon M\to N$ continuous iff $f(\overline{X})\subset\overline{f(X)}$ [duplicate]

Possible Duplicate: Continuity and Closure $f\colon M\to N$ is continuous iff for all $X\subset M$ we have that $f\left(\overline{X}\right)\subset\overline{f(X)}$. I only proved $\implies$. If $...
2
votes
1answer
575 views

Proving that $f(\bar Z)\subset\overline {f(Z)}$ when $f$ is a continuous map [duplicate]

I'm trying to solve this question from my textbook: Let $f:X\rightarrow Y$ be a continuous map and let $Z \subset X$. Prove the inclusion $f(\bar Z)\subset\overline {f(Z)}$. Thanks in advance ...
1
vote
2answers
237 views

proof $f(\overline{A})\subseteq \overline{f(A)} \Leftrightarrow f$ continuous [duplicate]

It'd be great if someone checked the proof I did for the following problem: $f:X\longrightarrow Y$, $f(\overline{A})\subseteq \overline{f(A)},\forall A\subseteq X \Leftrightarrow f$ continuous ...
0
votes
1answer
233 views

Continuous function between two topological spaces: an ELEMENTARY property. [duplicate]

I'm reading the first chapter of a book on general topology. It has a lot of small, simple exercises on almost all pages and I try to do them all to fully understand the subject. This one I did not ...
0
votes
1answer
47 views

Basic Topological problem on closed sets [duplicate]

How can one prove this proposition: proposition 5.15: $f$ is continuous iff $f(\overline{A}) \subset \overline{f(A)}$, where $\overline{A}$ denotes the closure of a set $A.$ It seems to be a small ...
0
votes
0answers
58 views

For $A \subset X$ prove that $ f(\overline{A}) \subset \overline{f(A)}$ [duplicate]

Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces and let $f: X \to Y $ be a continuous function. For $A \subset X$ prove that $$ f(\overline{A}) \subset \overline{f(A)}$$ Definition: A mapping $f:...
0
votes
1answer
49 views

Continuity and adherence in topology [duplicate]

Do we have $f:(E,\tau)\to(F,\sigma)$ continuous if and only if $\forall A\subset E, f(\overline{A})= \overline{f(A)}$ or just: $f:(E,\tau)\to(F,\sigma)$ continuous if and only if $$\forall A\...
12
votes
8answers
2k views

Why is the topological definition of continuous the way it is?

I was learning the definition of continuous as: $f\colon X\to Y$ is continuous if $f^{-1}(U)$ is open for every open $U\subseteq Y$ For me this translates to the following implication: IF $U \...
1
vote
3answers
2k views

$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$.

Definiton. $f$ is continuous at $a$ provided that for each open set $V$ in $Y$ containing $f(a)$ there is an open set $U$ in $X$ containing $a$ such that $f(U) \subset V$. Problem. $f$ is ...
5
votes
2answers
984 views

Show that $f(\bar A) \subset \overline{f(A)}$.

Let $X$ be a metric space, and $Y\subset X$ a subset. A point $x\in X$ is adherent to $Y$ if $B(x;r) \cap Y \neq \emptyset$ $\forall r > 0.$ The closure of $Y$ is then defined as $\bar Y := \{x\in ...
4
votes
3answers
209 views

Topology: Homeomorphism between finite complement topology in $\mathbb{R}$ and one of its subspaces

My class notes say that because $U=\mathbb{R}\backslash\{x_1,x_2,..,x_n\}$ has the same cardinality than $\mathbb{R}$, there exists a homeomorphism between: $(U,T_{cof})$ and $(\mathbb{R},T_{cof})$, ...
1
vote
1answer
736 views

Continuity and interior

I have questions about the relation between continuity and interior based on the article ;A map is continuous if and only if for every set, the image of closure is contained in the closure of image ...
1
vote
2answers
331 views

Prove that $f(x) = 0$ for all $x ∈ R$. [duplicate]

Let $f(x)$ be a continuous function such that $f(r) = 0$ for all rational numbers r. Prove that $f(x) = 0$ for all $x ∈ R$.
3
votes
4answers
114 views

Example of a continuous function s.t. $f(\overline{A}) \subsetneq \overline{f(A)}$

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 ...
4
votes
3answers
201 views

How can a continuous function induce a proper inclusion $f(\overline{A})\subsetneq \overline{f(A)}$?

Let $f:(X, d_X)\longrightarrow (Y, d_Y)$ be a continuous function between two metric spaces, $A\subseteq X$. We have $f(\overline{A})\subseteq \overline{f(A)}$ from this question. Can you please ...

15 30 50 per page