Definition & Concept Mastery Of Continous Images Quiz

1) If f:X→Y is continuous and X is connected, then f(X) must also be connected.

True

False

True, because the continuous image of a connected set must remain connected; a continuous function cannot create gaps.

Explanation

True, because the continuous image of a connected set must remain connected; a continuous function cannot create gaps.

2) Why do continuous images preserve connectedness?

Continuous functions do not break apart intervals.

Continuous functions map open sets to open sets.

Continuous functions preserve closures of all sets.

Continuous functions preserve compactness.

Continuous functions cannot introduce jumps or tears, so a connected set stays connected.

Explanation

Continuous functions cannot introduce jumps or tears, so a connected set stays connected.

3) Let X = [0,1] and f(x) = sin(x). What is true about f(X)?

F(X) is disconnected

F(X) is connected

F(X) has exactly two components

F(X) is not an interval

The image of a connected interval under a continuous function is always a connected interval, here [0, sin(1)].

Explanation

The image of a connected interval under a continuous function is always a connected interval, here [0, sin(1)].

4) A continuous image of a disconnected space must also be disconnected.

True

False

False, because a disconnected domain can still map to a connected set. For example, {-1,1} mapped by f(x)=0 gives a connected single-point image.

Explanation

False, because a disconnected domain can still map to a connected set. For example, {-1,1} mapped by f(x)=0 gives a connected single-point image.

5) Which condition is necessary for continuous images of connected sets to be connected?

Domain must be compact

Function must be injective

Function must be continuous

Codomain must be Hausdorff

Only continuity is required; compactness, injectivity, and Hausdorffness are irrelevant.

Explanation

Only continuity is required; compactness, injectivity, and Hausdorffness are irrelevant.

6) Select all functions that are continuous (on domain ℝ).

F (x) =

F(x)=⌊x⌋

F(x)=

F(x) = sin(x)

A function is continuous if it can be drawn without lifting a pencil from the paper. The function f(x) = [x] is discontinuous at x = 0, making it not continuous on the entire real line. The function f(x)= ${x}^{2}$ is continuous everywhere, f (x) = ${e}^{x}$ is a polynomial and hence continuous for all x, and f(x) = sin(x) is continuous as well due to the nature of trigonometric functions.

Explanation

A function is continuous if it can be drawn without lifting a pencil from the paper. The function f(x) = [x] is discontinuous at x = 0, making it not continuous on the entire real line. The function f(x)= ${x}^{2}$ is continuous everywhere, f (x) = ${e}^{x}$ is a polynomial and hence continuous for all x, and f(x) = sin(x) is continuous as well due to the nature of trigonometric functions.

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7) If X is connected and f:X→Y is continuous, what must be true about f(X)?

It is either empty or has one component

It must be compact

It must be both closed and open

It must contain all limit points of X

A continuous image of a connected space cannot split into multiple pieces.

Explanation

A continuous image of a connected space cannot split into multiple pieces.

8) If f(X) is disconnected, then X must be disconnected.

True

False

False, because a disconnected image does not imply the domain was disconnected; the function may simply not be continuous.

Explanation

False, because a disconnected image does not imply the domain was disconnected; the function may simply not be continuous.

9) Which counterexample shows that a disconnected domain can have a connected image?

X = {-1,1}, f(x) = 0

X = (0,1) ∪ (2,3), f(x) = x

X = {0} ∪ {1}, f(x) = x

X = {0} ∪ {1}, f(x) = |x−0.5|

{-1,1} is disconnected, but f(x)=0 maps both points to a single point, producing a connected image.

Explanation

{-1,1} is disconnected, but f(x)=0 maps both points to a single point, producing a connected image.

10) Let X= ℝ & f:ℝ→ℝ be continuous. Which must be connected?

F(ℝ)

ℝ \ f(ℝ)

Each preimage f−1(y)

All open subsets of ℝ

The continuous image of the connected real line must itself be connected.

Explanation

The continuous image of the connected real line must itself be connected.

11) A continuous image of a connected set in ℝ² must always be an interval.

True

False

The statement is false because a continuous image of a connected set in ℝ ² does not always result in an interval. Connected sets can map to more complex structures that are not simply intervals, such as curves or surfaces, which may not maintain the interval property.

Explanation

The statement is false because a continuous image of a connected set in ℝ ² does not always result in an interval. Connected sets can map to more complex structures that are not simply intervals, such as curves or surfaces, which may not maintain the interval property.

12) Let f:[0,2]→ℝ be continuous. Which can describe f([0,2])?

A single point

A nonempty finite set

A union of two separated intervals

One interval of any shape (open, closed, half-open)

The image of a connected interval must be an interval, possibly a single point or any type of interval.

Explanation

The image of a connected interval must be an interval, possibly a single point or any type of interval.

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13) Which statements about continuous functions and connectedness are true?

Continuous maps preserve connectedness

A constant function is always continuous

If X is connected and f(X) is disconnected, continuity fails

Connectedness is preserved by any function

Continuous maps always preserve connectedness, and constant functions are continuous.

Explanation

Continuous maps always preserve connectedness, and constant functions are continuous.

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14) Let f:X→Y be continuous. What could cause f(X) to fail to be connected?

X is not connected

F is not injective

Y is not compact

F is constant

If the domain is disconnected, the image may also be disconnected. Injectivity, compactness, and constant functions do not break connectedness.

Explanation

If the domain is disconnected, the image may also be disconnected. Injectivity, compactness, and constant functions do not break connectedness.

15) If f:X→Y is continuous and surjective, and Y is connected, then X must also be connected.

True

False

False, because a disconnected domain can still map onto a connected space. Example: {-1,1} → {0}.

Explanation

False, because a disconnected domain can still map onto a connected space. Example: {-1,1} → {0}.

Definition & Concept Mastery of Continous Images Quiz