Is Continuity Defined by the Behavior of Function Over Closure of Sets?

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Homework Help Overview

The discussion revolves around the concept of continuity in the context of functions from \(\mathbb{R}^n\) to \(\mathbb{R}^m\). The original poster seeks guidance on proving that a function is continuous if and only if the image of the closure of a set is contained within the closure of the image of that set.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various definitions of continuity and closure, with some suggesting the equivalence of different definitions. Questions are raised about how to connect the definition of continuity with the properties of closure, particularly in relation to sequences and limit points.

Discussion Status

The discussion is active, with participants exploring different definitions and their implications. Some guidance has been offered regarding the relationship between sequences and continuity, as well as the role of closure in this context. There is no explicit consensus yet, but productive lines of reasoning are being developed.

Contextual Notes

Participants note that the definitions of continuity and closure may vary, which could affect the approach to the problem. There is also mention of a hint involving the closure of a set of terms from a subsequence, indicating a specific direction for exploration.

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Hi!
Please, give me some guidance in solving this problem.

Let [itex]f:{\mathbb{R}}^n\longrightarrow{\mathbb{R}}^m[/itex].
Show that f is continuous iff for all [itex]M\subset{\mathbb{R}}^n[/itex] the inclusion [itex]f(closM)\subseteq{clos{}f(M)}[/itex] holds.(closM denotes the closure of the set M)

Please, ask me some guiding questions to lead me in the right direction straightaway. :smile:
 
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First off you should tell us what your definition of continuous is so that we do not use one of the many equivalent conditions that we aren't supposed to (eg inverse image of open set is open).
 
The definition we mainly use is:

The mapping [itex]f:X\longrightarrow{Y}[/itex] is continuous in [itex]x_0\in{X}[/itex] iff for all [itex]\epsilon>0[/itex] there exists [itex]\delta>0[/itex] such that for all [itex]x\in{X}[/itex] with[itex]d_X(x,x_0)<\delta[/itex] the following inequality holds: [itex]d_Y(f(x),f(x_0))<\epsilon[/itex]

I forgot to tell about the hint. It says we should use the closure of the set of the terms of some partial sequence, which should then be mapped by f. I can't get anything out of it.
I found an alternative proof of the above proposition (in Djedonne's "Analysis"), but it uses adherent points of M...which I don't think we are supposed to use, because we didn't explicitly treated these concepts.
 
So, you're specifically talking about this as a metric topology.

Do you know the definition that a function is continuous at x if and only if for all sequences x_n tending to x then f(x_n) tends to x. You shuold try to prove this is equivalent to your commonly used definition of continuity.
 
OK, but will it bring closure with it into the game?
 
Again, how easily useful this hint is will depend upon how *you* define the closure. There are at least 4 different (but equivalent) ways of doing this.
 
Our definition of closure is:
The closure of a set is the union of this set with all its limit points.

How can I formally connect the definition of continuity with the above def. of closure to get a statement about some terms of a subsequence in X converging to some x_0? What should happen when the closure of the set of these terms (hint) is mapped under f? Should there be a contradiction?
 
Let me retell you what you already know.

Let x_n be a series in M that tend to x, ie a point in the closure.

What happnens when we apply f?

Now, every point in the closure of M is the limit of such a sequence.

Hence the result is true.
 

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